def test_prioritised_transactions(self):
# Ensure that fee deltas used via prioritisetransaction are
# correctly used by replacement logic
# 1. Check that feeperkb uses modified fees
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
# Higher fee, but the actual fee per KB is much lower.
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.001*COIN), CScript([b'a'*740000]))]
tx1b_hex = txToHex(tx1b)
# Verify tx1b cannot replace tx1a.
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx1b_hex, True)
# Use prioritisetransaction to set tx1a's fee to 0.
self.nodes[0].prioritisetransaction(txid=tx1a_txid, fee_delta=int(-0.1*COIN))
# Now tx1b should be able to replace tx1a
tx1b_txid = self.nodes[0].sendrawtransaction(tx1b_hex, True)
assert(tx1b_txid in self.nodes[0].getrawmempool())
# 2. Check that absolute fee checks use modified fee.
tx1_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
tx2a = CTransaction()
tx2a.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx2a_hex = txToHex(tx2a)
self.nodes[0].sendrawtransaction(tx2a_hex, True)
# Lower fee, but we'll prioritise it
tx2b = CTransaction()
tx2b.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2b.vout = [CTxOut(int(1.01 * COIN), CScript([b'a' * 35]))]
tx2b.rehash()
tx2b_hex = txToHex(tx2b)
# Verify tx2b cannot replace tx2a.
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx2b_hex, True)
# Now prioritise tx2b to have a higher modified fee
self.nodes[0].prioritisetransaction(txid=tx2b.hash, fee_delta=int(0.1*COIN))
# tx2b should now be accepted
tx2b_txid = self.nodes[0].sendrawtransaction(tx2b_hex, True)
assert(tx2b_txid in self.nodes[0].getrawmempool())
def test_too_many_replacements(self):
"""Replacements that evict too many transactions are rejected"""
# Try directly replacing more than MAX_REPLACEMENT_LIMIT
# transactions
# Start by creating a single transaction with many outputs
initial_nValue = 10*COIN
utxo = make_utxo(self.nodes[0], initial_nValue)
fee = int(0.0001*COIN)
split_value = int((initial_nValue-fee)/(MAX_REPLACEMENT_LIMIT+1))
outputs = []
for i in range(MAX_REPLACEMENT_LIMIT+1):
outputs.append(CTxOut(split_value, CScript([1])))
splitting_tx = CTransaction()
splitting_tx.vin = [CTxIn(utxo, nSequence=0)]
splitting_tx.vout = outputs
splitting_tx_hex = txToHex(splitting_tx)
txid = self.nodes[0].sendrawtransaction(splitting_tx_hex, True)
txid = int(txid, 16)
# Now spend each of those outputs individually
for i in range(MAX_REPLACEMENT_LIMIT+1):
tx_i = CTransaction()
tx_i.vin = [CTxIn(COutPoint(txid, i), nSequence=0)]
tx_i.vout = [CTxOut(split_value - fee, CScript([b'a' * 35]))]
tx_i_hex = txToHex(tx_i)
self.nodes[0].sendrawtransaction(tx_i_hex, True)
# Now create doublespend of the whole lot; should fail.
# Need a big enough fee to cover all spending transactions and have
# a higher fee rate
double_spend_value = (split_value-100*fee)*(MAX_REPLACEMENT_LIMIT+1)
inputs = []
for i in range(MAX_REPLACEMENT_LIMIT+1):
inputs.append(CTxIn(COutPoint(txid, i), nSequence=0))
double_tx = CTransaction()
double_tx.vin = inputs
double_tx.vout = [CTxOut(double_spend_value, CScript([b'a']))]
double_tx_hex = txToHex(double_tx)
# This will raise an exception
assert_raises_rpc_error(-26, "too many potential replacements", self.nodes[0].sendrawtransaction, double_tx_hex, True)
# If we remove an input, it should pass
double_tx = CTransaction()
double_tx.vin = inputs[0:-1]
double_tx.vout = [CTxOut(double_spend_value, CScript([b'a']))]
double_tx_hex = txToHex(double_tx)
self.nodes[0].sendrawtransaction(double_tx_hex, True)
def test_simple_doublespend(self):
"""Simple doublespend"""
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
# make_utxo may have generated a bunch of blocks, so we need to sync
# before we can spend the coins generated, or else the resulting
# transactions might not be accepted by our peers.
self.sync_all()
feeout = CTxOut(int(0.1*COIN), CScript())
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35])), feeout]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
self.sync_all()
# Should fail because we haven't changed the fee
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(1 * COIN, CScript([b'b' * 35])), feeout]
tx1b_hex = txToHex(tx1b)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx1b_hex, True)
# This will raise an exception due to transaction replacement being disabled
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[1].sendrawtransaction, tx1b_hex, True)
# Extra 0.1 BTC fee
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.9 * COIN), CScript([b'b' * 35])), feeout, feeout]
tx1b_hex = txToHex(tx1b)
# Replacement still disabled even with "enough fee"
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[1].sendrawtransaction, tx1b_hex, True)
# Works when enabled
tx1b_txid = self.nodes[0].sendrawtransaction(tx1b_hex, True)
mempool = self.nodes[0].getrawmempool()
assert (tx1a_txid not in mempool)
assert (tx1b_txid in mempool)
assert_equal(tx1b_hex, self.nodes[0].getrawtransaction(tx1b_txid))
# Second node is running mempoolreplacement=0, will not replace originally-seen txn
mempool = self.nodes[1].getrawmempool()
assert tx1a_txid in mempool
assert tx1b_txid not in mempool
def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.0001*COIN, _total_txs=None):
if _total_txs is None:
_total_txs = [0]
if _total_txs[0] >= max_txs:
return
txout_value = (initial_value - fee) // tree_width
if txout_value < fee:
return
vout = [CTxOut(txout_value, CScript([i+1]))
for i in range(tree_width)]
tx = CTransaction()
tx.vin = [CTxIn(prevout, nSequence=0)]
tx.vout = vout
tx_hex = txToHex(tx)
assert(len(tx.serialize()) < 100000)
txid = self.nodes[0].sendrawtransaction(tx_hex, True)
yield tx
_total_txs[0] += 1
txid = int(txid, 16)
for i, txout in enumerate(tx.vout):
for x in branch(COutPoint(txid, i), txout_value,
max_txs,
tree_width=tree_width, fee=fee,
_total_txs=_total_txs):
yield x
def test_new_unconfirmed_inputs(self):
"""Replacements that add new unconfirmed inputs are rejected"""
confirmed_utxo = make_utxo(self.nodes[0], int(1.1*COIN))
unconfirmed_utxo = make_utxo(self.nodes[0], int(0.1*COIN), False)
tx1 = CTransaction()
tx1.vin = [CTxIn(confirmed_utxo)]
tx1.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx1_hex = txToHex(tx1)
self.nodes[0].sendrawtransaction(tx1_hex, True)
tx2 = CTransaction()
tx2.vin = [CTxIn(confirmed_utxo), CTxIn(unconfirmed_utxo)]
tx2.vout = tx1.vout
tx2_hex = txToHex(tx2)
# This will raise an exception
assert_raises_rpc_error(-26, "replacement-adds-unconfirmed", self.nodes[0].sendrawtransaction, tx2_hex, True)
def test_bip68_not_consensus(self):
assert(get_bip9_status(self.nodes[0], 'csv')['status'] != 'active')
txid = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), 2)
tx1 = FromHex(CTransaction(), self.nodes[0].getrawtransaction(txid))
tx1.rehash()
# Make an anyone-can-spend transaction
tx2 = CTransaction()
tx2.nVersion = 1
tx2.vin = [CTxIn(COutPoint(tx1.sha256, 0), nSequence=0)]
tx2.vout = [CTxOut(int(tx1.vout[0].nValue - self.relayfee*COIN), CScript([b'a']))]
# sign tx2
tx2_raw = self.nodes[0].signrawtransactionwithwallet(ToHex(tx2))["hex"]
tx2 = FromHex(tx2, tx2_raw)
tx2.rehash()
self.nodes[0].sendrawtransaction(ToHex(tx2))
# Now make an invalid spend of tx2 according to BIP68
sequence_value = 100 # 100 block relative locktime
tx3 = CTransaction()
tx3.nVersion = 2
tx3.vin = [CTxIn(COutPoint(tx2.sha256, 0), nSequence=sequence_value)]
tx3.vout = [CTxOut(int(tx2.vout[0].nValue - self.relayfee * COIN), CScript([b'a' * 35]))]
tx3.rehash()
assert_raises_rpc_error(-26, NOT_FINAL_ERROR, self.nodes[0].sendrawtransaction, ToHex(tx3))
# make a block that violates bip68; ensure that the tip updates
tip = int(self.nodes[0].getbestblockhash(), 16)
block = create_block(tip, create_coinbase(self.nodes[0].getblockcount()+1))
block.nVersion = 3
block.vtx.extend([tx1, tx2, tx3])
block.hashMerkleRoot = block.calc_merkle_root()
block.rehash()
add_witness_commitment(block)
block.solve()
self.nodes[0].submitblock(bytes_to_hex_str(block.serialize(True)))
assert_equal(self.nodes[0].getbestblockhash(), block.hash)
def test_replacement_feeperkb(self):
"""Replacement requires fee-per-KB to be higher"""
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx1a_hex = txToHex(tx1a)
self.nodes[0].sendrawtransaction(tx1a_hex, True)
# Higher fee, but the fee per KB is much lower, so the replacement is
# rejected.
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.001*COIN), CScript([b'a'*999000]))]
tx1b_hex = txToHex(tx1b)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, tx1b_hex, True)
def test_disable_flag(self):
# Create some unconfirmed inputs
new_addr = self.nodes[0].getnewaddress()
self.nodes[0].sendtoaddress(new_addr, 2) # send 2 BTC
utxos = self.nodes[0].listunspent(0, 0)
assert len(utxos) > 0
utxo = utxos[0]
tx1 = CTransaction()
value = int(satoshi_round(utxo["amount"] - self.relayfee)*COIN)
# Check that the disable flag disables relative locktime.
# If sequence locks were used, this would require 1 block for the
# input to mature.
sequence_value = SEQUENCE_LOCKTIME_DISABLE_FLAG | 1
tx1.vin = [CTxIn(COutPoint(int(utxo["txid"], 16), utxo["vout"]), nSequence=sequence_value)]
tx1.vout = [CTxOut(value, CScript([b'a']))]
tx1_signed = self.nodes[0].signrawtransactionwithwallet(ToHex(tx1))["hex"]
tx1_id = self.nodes[0].sendrawtransaction(tx1_signed)
tx1_id = int(tx1_id, 16)
# This transaction will enable sequence-locks, so this transaction should
# fail
tx2 = CTransaction()
tx2.nVersion = 2
sequence_value = sequence_value & 0x7fffffff
tx2.vin = [CTxIn(COutPoint(tx1_id, 0), nSequence=sequence_value)]
tx2.vout = [CTxOut(int(value - self.relayfee * COIN), CScript([b'a' * 35]))]
tx2.rehash()
assert_raises_rpc_error(-26, NOT_FINAL_ERROR, self.nodes[0].sendrawtransaction, ToHex(tx2))
# Setting the version back down to 1 should disable the sequence lock,
# so this should be accepted.
tx2.nVersion = 1
self.nodes[0].sendrawtransaction(ToHex(tx2))
def test_doublespend_chain(self):
"""Doublespend of a long chain"""
initial_nValue = 50*COIN
tx0_outpoint = make_utxo(self.nodes[0], initial_nValue)
prevout = tx0_outpoint
remaining_value = initial_nValue
chain_txids = []
while remaining_value > 10*COIN:
remaining_value -= 1*COIN
tx = CTransaction()
tx.vin = [CTxIn(prevout, nSequence=0)]
feeout = CTxOut(1*COIN)
tx.vout = [CTxOut(remaining_value, CScript([1, OP_DROP] * 15 + [1])), feeout]
tx_hex = txToHex(tx)
txid = self.nodes[0].sendrawtransaction(tx_hex, True)
chain_txids.append(txid)
prevout = COutPoint(int(txid, 16), 0)
# Whether the double-spend is allowed is evaluated by including all
# child fees - 40 BTC - so this attempt is rejected.
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - 30 * COIN, CScript([1] * 35)), CTxOut(30*COIN)]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, dbl_tx_hex, True)
# Accepted with sufficient fee
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(1 * COIN, CScript([1] * 35)), CTxOut(49*COIN)]
dbl_tx_hex = txToHex(dbl_tx)
self.nodes[0].sendrawtransaction(dbl_tx_hex, True)
mempool = self.nodes[0].getrawmempool()
for doublespent_txid in chain_txids:
assert(doublespent_txid not in mempool)
def test_spends_of_conflicting_outputs(self):
"""Replacements that spend conflicting tx outputs are rejected"""
utxo1 = make_utxo(self.nodes[0], int(1.2*COIN))
utxo2 = make_utxo(self.nodes[0], 3*COIN)
tx1a = CTransaction()
tx1a.vin = [CTxIn(utxo1, nSequence=0)]
tx1a.vout = [CTxOut(int(1.1 * COIN), CScript([b'a' * 35]))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
tx1a_txid = int(tx1a_txid, 16)
# Direct spend an output of the transaction we're replacing.
tx2 = CTransaction()
tx2.vin = [CTxIn(utxo1, nSequence=0), CTxIn(utxo2, nSequence=0)]
tx2.vin.append(CTxIn(COutPoint(tx1a_txid, 0), nSequence=0))
tx2.vout = tx1a.vout
tx2_hex = txToHex(tx2)
# This will raise an exception
assert_raises_rpc_error(-26, "bad-txns-spends-conflicting-tx", self.nodes[0].sendrawtransaction, tx2_hex, True)
# Spend tx1a's output to test the indirect case.
tx1b = CTransaction()
tx1b.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)]
tx1b.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx1b_hex = txToHex(tx1b)
tx1b_txid = self.nodes[0].sendrawtransaction(tx1b_hex, True)
tx1b_txid = int(tx1b_txid, 16)
tx2 = CTransaction()
tx2.vin = [CTxIn(utxo1, nSequence=0), CTxIn(utxo2, nSequence=0),
CTxIn(COutPoint(tx1b_txid, 0))]
tx2.vout = tx1a.vout
tx2_hex = txToHex(tx2)
# This will raise an exception
assert_raises_rpc_error(-26, "bad-txns-spends-conflicting-tx", self.nodes[0].sendrawtransaction, tx2_hex, True)
def make_utxo(node, amount, confirmed=True, scriptPubKey=CScript([1])):
"""Create a txout with a given amount and scriptPubKey
Mines coins as needed.
confirmed - txouts created will be confirmed in the blockchain;
unconfirmed otherwise.
"""
fee = 1*COIN
while node.getbalance()['bitcoin'] < satoshi_round((amount + fee)/COIN):
node.generate(100)
new_addr = node.getnewaddress()
unblinded_addr = node.validateaddress(new_addr)["unconfidential"]
txidstr = node.sendtoaddress(new_addr, satoshi_round((amount+fee)/COIN))
tx1 = node.getrawtransaction(txidstr, 1)
txid = int(txidstr, 16)
i = None
for i, txout in enumerate(tx1['vout']):
if txout['scriptPubKey']['type'] == "fee":
continue # skip fee outputs
if txout['scriptPubKey']['addresses'] == [unblinded_addr]:
break
assert i is not None
tx2 = CTransaction()
tx2.vin = [CTxIn(COutPoint(txid, i))]
tx1raw = CTransaction()
tx1raw.deserialize(BytesIO(hex_str_to_bytes(node.getrawtransaction(txidstr))))
feeout = CTxOut(CTxOutValue(tx1raw.vout[i].nValue.getAmount() - amount))
tx2.vout = [CTxOut(amount, scriptPubKey), feeout]
tx2.rehash()
signed_tx = node.signrawtransactionwithwallet(txToHex(tx2))
txid = node.sendrawtransaction(signed_tx['hex'], True)
# If requested, ensure txouts are confirmed.
if confirmed:
mempool_size = len(node.getrawmempool())
while mempool_size > 0:
node.generate(1)
new_size = len(node.getrawmempool())
# Error out if we have something stuck in the mempool, as this
# would likely be a bug.
assert(new_size < mempool_size)
mempool_size = new_size
return COutPoint(int(txid, 16), 0)
def make_utxo(node, amount, confirmed=True, scriptPubKey=CScript([1])):
"""Create a txout with a given amount and scriptPubKey
Mines coins as needed.
confirmed - txouts created will be confirmed in the blockchain;
unconfirmed otherwise.
"""
fee = 1*COIN
while node.getbalance() < satoshi_round((amount + fee)/COIN):
node.generate(100)
new_addr = node.getnewaddress()
txid = node.sendtoaddress(new_addr, satoshi_round((amount+fee)/COIN))
tx1 = node.getrawtransaction(txid, 1)
txid = int(txid, 16)
i = None
for i, txout in enumerate(tx1['vout']):
if txout['scriptPubKey']['addresses'] == [new_addr]:
break
assert i is not None
tx2 = CTransaction()
tx2.vin = [CTxIn(COutPoint(txid, i))]
tx2.vout = [CTxOut(amount, scriptPubKey)]
tx2.rehash()
signed_tx = node.signrawtransactionwithwallet(txToHex(tx2))
txid = node.sendrawtransaction(signed_tx['hex'], True)
# If requested, ensure txouts are confirmed.
if confirmed:
mempool_size = len(node.getrawmempool())
while mempool_size > 0:
node.generate(1)
new_size = len(node.getrawmempool())
# Error out if we have something stuck in the mempool, as this
# would likely be a bug.
assert(new_size < mempool_size)
mempool_size = new_size
return COutPoint(int(txid, 16), 0)
def test_nonzero_locks(orig_tx, node, relayfee, use_height_lock):
sequence_value = 1
if not use_height_lock:
sequence_value |= SEQUENCE_LOCKTIME_TYPE_FLAG
tx = CTransaction()
tx.nVersion = 2
tx.vin = [CTxIn(COutPoint(orig_tx.sha256, 0), nSequence=sequence_value)]
tx.vout = [CTxOut(int(orig_tx.vout[0].nValue - relayfee * COIN), CScript([b'a' * 35]))]
tx.rehash()
if (orig_tx.hash in node.getrawmempool()):
# sendrawtransaction should fail if the tx is in the mempool
assert_raises_rpc_error(-26, NOT_FINAL_ERROR, node.sendrawtransaction, ToHex(tx))
else:
# sendrawtransaction should succeed if the tx is not in the mempool
node.sendrawtransaction(ToHex(tx))
return tx
def test_opt_in(self):
"""Replacing should only work if orig tx opted in"""
tx0_outpoint = make_utxo(self.nodes[0], int(1.1 * COIN))
# Create a non-opting in transaction
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0xffffffff)]
tx1a.vout = [CTxOut(1 * COIN, DUMMY_P2WPKH_SCRIPT)]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, 0)
# This transaction isn't shown as replaceable
assert_equal(
self.nodes[0].getmempoolentry(tx1a_txid)['bip125-replaceable'],
False)
# Shouldn't be able to double-spend
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.9 * COIN), DUMMY_P2WPKH_SCRIPT)]
tx1b_hex = txToHex(tx1b)
# This will raise an exception
assert_raises_rpc_error(-26, "txn-mempool-conflict",
self.nodes[0].sendrawtransaction, tx1b_hex, 0)
tx1_outpoint = make_utxo(self.nodes[0], int(1.1 * COIN))
# Create a different non-opting in transaction
tx2a = CTransaction()
tx2a.vin = [CTxIn(tx1_outpoint, nSequence=0xfffffffe)]
tx2a.vout = [CTxOut(1 * COIN, DUMMY_P2WPKH_SCRIPT)]
tx2a_hex = txToHex(tx2a)
tx2a_txid = self.nodes[0].sendrawtransaction(tx2a_hex, 0)
# Still shouldn't be able to double-spend
tx2b = CTransaction()
tx2b.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2b.vout = [CTxOut(int(0.9 * COIN), DUMMY_P2WPKH_SCRIPT)]
tx2b_hex = txToHex(tx2b)
# This will raise an exception
assert_raises_rpc_error(-26, "txn-mempool-conflict",
self.nodes[0].sendrawtransaction, tx2b_hex, 0)
# Now create a new transaction that spends from tx1a and tx2a
# opt-in on one of the inputs
# Transaction should be replaceable on either input
tx1a_txid = int(tx1a_txid, 16)
tx2a_txid = int(tx2a_txid, 16)
tx3a = CTransaction()
tx3a.vin = [
CTxIn(COutPoint(tx1a_txid, 0), nSequence=0xffffffff),
CTxIn(COutPoint(tx2a_txid, 0), nSequence=0xfffffffd)
]
tx3a.vout = [
CTxOut(int(0.9 * COIN), CScript([b'c'])),
CTxOut(int(0.9 * COIN), CScript([b'd']))
]
tx3a_hex = txToHex(tx3a)
tx3a_txid = self.nodes[0].sendrawtransaction(tx3a_hex, 0)
# This transaction is shown as replaceable
assert_equal(
self.nodes[0].getmempoolentry(tx3a_txid)['bip125-replaceable'],
True)
tx3b = CTransaction()
tx3b.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)]
tx3b.vout = [CTxOut(int(0.5 * COIN), DUMMY_P2WPKH_SCRIPT)]
tx3b_hex = txToHex(tx3b)
tx3c = CTransaction()
tx3c.vin = [CTxIn(COutPoint(tx2a_txid, 0), nSequence=0)]
tx3c.vout = [CTxOut(int(0.5 * COIN), DUMMY_P2WPKH_SCRIPT)]
tx3c_hex = txToHex(tx3c)
self.nodes[0].sendrawtransaction(tx3b_hex, 0)
# If tx3b was accepted, tx3c won't look like a replacement,
# but make sure it is accepted anyway
self.nodes[0].sendrawtransaction(tx3c_hex, 0)
def test_doublespend_tree(self):
"""Doublespend of a big tree of transactions"""
initial_nValue = 50*COIN
tx0_outpoint = make_utxo(self.nodes[0], initial_nValue)
def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.0001*COIN, _total_txs=None):
if _total_txs is None:
_total_txs = [0]
if _total_txs[0] >= max_txs:
return
txout_value = (initial_value - fee) // tree_width
if txout_value < fee:
return
vout = [CTxOut(txout_value, CScript([i+1]))
for i in range(tree_width)]
tx = CTransaction()
tx.vin = [CTxIn(prevout, nSequence=0)]
tx.vout = vout
tx_hex = txToHex(tx)
assert(len(tx.serialize()) < 100000)
txid = self.nodes[0].sendrawtransaction(tx_hex, True)
yield tx
_total_txs[0] += 1
txid = int(txid, 16)
for i, txout in enumerate(tx.vout):
for x in branch(COutPoint(txid, i), txout_value,
max_txs,
tree_width=tree_width, fee=fee,
_total_txs=_total_txs):
yield x
fee = int(0.0001*COIN)
n = MAX_REPLACEMENT_LIMIT
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)
# Attempt double-spend, will fail because too little fee paid
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - fee * n, CScript([1] * 35))]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, dbl_tx_hex, True)
# 1 BTC fee is enough
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - fee * n - 1 * COIN, CScript([1] * 35))]
dbl_tx_hex = txToHex(dbl_tx)
self.nodes[0].sendrawtransaction(dbl_tx_hex, True)
mempool = self.nodes[0].getrawmempool()
for tx in tree_txs:
tx.rehash()
assert (tx.hash not in mempool)
# Try again, but with more total transactions than the "max txs
# double-spent at once" anti-DoS limit.
for n in (MAX_REPLACEMENT_LIMIT+1, MAX_REPLACEMENT_LIMIT*2):
fee = int(0.0001*COIN)
tx0_outpoint = make_utxo(self.nodes[0], initial_nValue)
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - 2 * fee * n, CScript([1] * 35))]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception
assert_raises_rpc_error(-26, "too many potential replacements", self.nodes[0].sendrawtransaction, dbl_tx_hex, True)
for tx in tree_txs:
tx.rehash()
self.nodes[0].getrawtransaction(tx.hash)
def run_test(self):
node = self.nodes[0]
self.log.info('Start with empty mempool, and 200 blocks')
self.mempool_size = 0
assert_equal(node.getblockcount(), 200)
assert_equal(node.getmempoolinfo()['size'], self.mempool_size)
coins = node.listunspent()
self.log.info('Should not accept garbage to testmempoolaccept')
assert_raises_rpc_error(-3, 'Expected type array, got string', lambda: node.testmempoolaccept(rawtxs='ff00baar'))
assert_raises_rpc_error(-8, 'Array must contain exactly one raw transaction for now', lambda: node.testmempoolaccept(rawtxs=['ff00baar', 'ff22']))
assert_raises_rpc_error(-22, 'TX decode failed', lambda: node.testmempoolaccept(rawtxs=['ff00baar']))
self.log.info('A transaction already in the blockchain')
coin = coins.pop() # Pick a random coin(base) to spend
raw_tx_in_block = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{'txid': coin['txid'], 'vout': coin['vout']}],
outputs=[{node.getnewaddress(): 0.3}, {node.getnewaddress(): 49}],
))['hex']
txid_in_block = node.sendrawtransaction(hexstring=raw_tx_in_block, maxfeerate=0)
node.generate(1)
self.mempool_size = 0
self.check_mempool_result(
result_expected=[{'txid': txid_in_block, 'allowed': False, 'reject-reason': '18: txn-already-known'}],
rawtxs=[raw_tx_in_block],
)
self.log.info('A transaction not in the mempool')
fee = 0.00000700
raw_tx_0 = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{"txid": txid_in_block, "vout": 0, "sequence": BIP125_SEQUENCE_NUMBER}], # RBF is used later
outputs=[{node.getnewaddress(): 0.3 - fee}],
))['hex']
tx = CTransaction()
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
txid_0 = tx.rehash()
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': True}],
rawtxs=[raw_tx_0],
)
self.log.info('A final transaction not in the mempool')
coin = coins.pop() # Pick a random coin(base) to spend
raw_tx_final = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{'txid': coin['txid'], 'vout': coin['vout'], "sequence": 0xffffffff}], # SEQUENCE_FINAL
outputs=[{node.getnewaddress(): 0.025}],
locktime=node.getblockcount() + 2000, # Can be anything
))['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_final)))
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': True}],
rawtxs=[tx.serialize().hex()],
maxfeerate=0,
)
node.sendrawtransaction(hexstring=raw_tx_final, maxfeerate=0)
self.mempool_size += 1
self.log.info('A transaction in the mempool')
node.sendrawtransaction(hexstring=raw_tx_0)
self.mempool_size += 1
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': False, 'reject-reason': '18: txn-already-in-mempool'}],
rawtxs=[raw_tx_0],
)
self.log.info('A transaction that replaces a mempool transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vout[0].nValue -= int(fee * COIN) # Double the fee
tx.vin[0].nSequence = BIP125_SEQUENCE_NUMBER + 1 # Now, opt out of RBF
raw_tx_0 = node.signrawtransactionwithwallet(tx.serialize().hex())['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
txid_0 = tx.rehash()
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': True}],
rawtxs=[raw_tx_0],
)
self.log.info('A transaction that conflicts with an unconfirmed tx')
# Send the transaction that replaces the mempool transaction and opts out of replaceability
node.sendrawtransaction(hexstring=tx.serialize().hex(), maxfeerate=0)
# take original raw_tx_0
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vout[0].nValue -= int(4 * fee * COIN) # Set more fee
# skip re-signing the tx
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '18: txn-mempool-conflict'}],
rawtxs=[tx.serialize().hex()],
maxfeerate=0,
)
self.log.info('A transaction with missing inputs, that never existed')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vin[0].prevout = COutPoint(hash=int('ff' * 32, 16), n=14)
# skip re-signing the tx
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'missing-inputs'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with missing inputs, that existed once in the past')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vin[0].prevout.n = 1 # Set vout to 1, to spend the other outpoint (49 coins) of the in-chain-tx we want to double spend
raw_tx_1 = node.signrawtransactionwithwallet(tx.serialize().hex())['hex']
txid_1 = node.sendrawtransaction(hexstring=raw_tx_1, maxfeerate=0)
# Now spend both to "clearly hide" the outputs, ie. remove the coins from the utxo set by spending them
raw_tx_spend_both = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[
{'txid': txid_0, 'vout': 0},
{'txid': txid_1, 'vout': 0},
],
outputs=[{node.getnewaddress(): 0.1}]
))['hex']
txid_spend_both = node.sendrawtransaction(hexstring=raw_tx_spend_both, maxfeerate=0)
node.generate(1)
self.mempool_size = 0
# Now see if we can add the coins back to the utxo set by sending the exact txs again
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': False, 'reject-reason': 'missing-inputs'}],
rawtxs=[raw_tx_0],
)
self.check_mempool_result(
result_expected=[{'txid': txid_1, 'allowed': False, 'reject-reason': 'missing-inputs'}],
rawtxs=[raw_tx_1],
)
self.log.info('Create a signed "reference" tx for later use')
raw_tx_reference = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{'txid': txid_spend_both, 'vout': 0}],
outputs=[{node.getnewaddress(): 0.05}],
))['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
# Reference tx should be valid on itself
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': True}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with no outputs')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout = []
# Skip re-signing the transaction for context independent checks from now on
# tx.deserialize(BytesIO(hex_str_to_bytes(node.signrawtransactionwithwallet(tx.serialize().hex())['hex'])))
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '16: bad-txns-vout-empty'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A really large transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin = [tx.vin[0]] * math.ceil(MAX_BLOCK_BASE_SIZE / len(tx.vin[0].serialize()))
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '16: bad-txns-oversize'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with negative output value')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].nValue *= -1
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '16: bad-txns-vout-negative'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with too large output value')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].nValue = 21000000 * COIN + 1
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '16: bad-txns-vout-toolarge'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with too large sum of output values')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout = [tx.vout[0]] * 2
tx.vout[0].nValue = 21000000 * COIN
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '16: bad-txns-txouttotal-toolarge'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with duplicate inputs')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin = [tx.vin[0]] * 2
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '16: bad-txns-inputs-duplicate'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A coinbase transaction')
# Pick the input of the first tx we signed, so it has to be a coinbase tx
raw_tx_coinbase_spent = node.getrawtransaction(txid=node.decoderawtransaction(hexstring=raw_tx_in_block)['vin'][0]['txid'])
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_coinbase_spent)))
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '16: coinbase'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('Some nonstandard transactions')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.nVersion = 3 # A version currently non-standard
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: version'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].scriptPubKey = CScript([OP_0]) # Some non-standard script
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: scriptpubkey'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].scriptSig = CScript([OP_HASH160]) # Some not-pushonly scriptSig
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: scriptsig-not-pushonly'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
output_p2sh_burn = CTxOut(nValue=540, scriptPubKey=CScript([OP_HASH160, hash160(b'burn'), OP_EQUAL]))
num_scripts = 100000 // len(output_p2sh_burn.serialize()) # Use enough outputs to make the tx too large for our policy
tx.vout = [output_p2sh_burn] * num_scripts
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: tx-size'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0] = output_p2sh_burn
tx.vout[0].nValue -= 1 # Make output smaller, such that it is dust for our policy
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: dust'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].scriptPubKey = CScript([OP_RETURN, b'\xff'])
tx.vout = [tx.vout[0]] * 2
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: multi-op-return'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A timelocked transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].nSequence -= 1 # Should be non-max, so locktime is not ignored
tx.nLockTime = node.getblockcount() + 1
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: non-final'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction that is locked by BIP68 sequence logic')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].nSequence = 2 # We could include it in the second block mined from now, but not the very next one
# Can skip re-signing the tx because of early rejection
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': '64: non-BIP68-final'}],
rawtxs=[tx.serialize().hex()],
maxfeerate=0,
)
def run_test(self):
[node] = self.nodes
node.add_p2p_connection(P2PDataStore())
# Get out of IBD
node.generatetoaddress(1, node.get_deterministic_priv_key().address)
tip = self.getbestblock(node)
self.log.info("Create some blocks with OP_1 coinbase for spending.")
blocks = []
for _ in range(20):
tip = self.build_block(tip)
blocks.append(tip)
node.p2p.send_blocks_and_test(blocks, node, success=True)
self.spendable_outputs = deque(block.vtx[0] for block in blocks)
self.log.info("Mature the blocks.")
node.generatetoaddress(100, node.get_deterministic_priv_key().address)
tip = self.getbestblock(node)
# To make compact and fast-to-verify transactions, we'll use
# CHECKDATASIG over and over with the same data.
# (Using the same stuff over and over again means we get to hit the
# node's signature cache and don't need to make new signatures every
# time.)
cds_message = b''
# r=1 and s=1 ecdsa, the minimum values.
cds_signature = bytes.fromhex('3006020101020101')
# Recovered pubkey
cds_pubkey = bytes.fromhex(
'03089b476b570d66fad5a20ae6188ebbaf793a4c2a228c65f3d79ee8111d56c932'
)
def minefunding2(n):
""" Mine a block with a bunch of outputs that are very dense
sigchecks when spent (2 sigchecks each); return the inputs that can
be used to spend. """
cds_scriptpubkey = CScript([
cds_message, cds_pubkey, OP_3DUP, OP_CHECKDATASIGVERIFY,
OP_CHECKDATASIGVERIFY
])
# The scriptsig is carefully padded to have size 26, which is the
# shortest allowed for 2 sigchecks for mempool admission.
# The resulting inputs have size 67 bytes, 33.5 bytes/sigcheck.
cds_scriptsig = CScript([b'x' * 16, cds_signature])
assert_equal(len(cds_scriptsig), 26)
self.log.debug(
"Gen {} with locking script {} unlocking script {} .".format(
n, cds_scriptpubkey.hex(), cds_scriptsig.hex()))
tx = self.spendable_outputs.popleft()
usable_inputs = []
txes = []
for _ in range(n):
tx = create_transaction(tx, cds_scriptpubkey)
txes.append(tx)
usable_inputs.append(
CTxIn(COutPoint(tx.sha256, 1), cds_scriptsig))
newtip = self.build_block(tip, txes)
node.p2p.send_blocks_and_test([newtip], node)
return usable_inputs, newtip
self.log.info("Funding special coins that have high sigchecks")
# mine 5000 funded outputs (10000 sigchecks)
# will be used pre-activation and post-activation
usable_inputs, tip = minefunding2(5000)
# assemble them into 50 txes with 100 inputs each (200 sigchecks)
submittxes_1 = []
while len(usable_inputs) >= 100:
tx = CTransaction()
tx.vin = [usable_inputs.pop() for _ in range(100)]
tx.vout = [CTxOut(0, CScript([OP_RETURN]))]
tx.rehash()
submittxes_1.append(tx)
# mine 5000 funded outputs (10000 sigchecks)
# will be used post-activation
usable_inputs, tip = minefunding2(5000)
# assemble them into 50 txes with 100 inputs each (200 sigchecks)
submittxes_2 = []
while len(usable_inputs) >= 100:
tx = CTransaction()
tx.vin = [usable_inputs.pop() for _ in range(100)]
tx.vout = [CTxOut(0, CScript([OP_RETURN]))]
tx.rehash()
submittxes_2.append(tx)
# Check high sigcheck transactions
self.log.info("Create transaction that have high sigchecks")
fundings = []
def make_spend(sigcheckcount):
# Add a funding tx to fundings, and return a tx spending that using
# scriptsig.
self.log.debug("Gen tx with {} sigchecks.".format(sigcheckcount))
def get_script_with_sigcheck(count):
return CScript([cds_message, cds_pubkey] +
(count - 1) * [OP_3DUP, OP_CHECKDATASIGVERIFY] +
[OP_CHECKDATASIG])
# get funds locked with OP_1
sourcetx = self.spendable_outputs.popleft()
# make funding that forwards to scriptpubkey
last_sigcheck_count = ((sigcheckcount - 1) % 30) + 1
fundtx = create_transaction(
sourcetx, get_script_with_sigcheck(last_sigcheck_count))
fill_sigcheck_script = get_script_with_sigcheck(30)
remaining_sigcheck = sigcheckcount
while remaining_sigcheck > 30:
fundtx.vout[0].nValue -= 1000
fundtx.vout.append(CTxOut(100, bytes(fill_sigcheck_script)))
remaining_sigcheck -= 30
fundtx.rehash()
fundings.append(fundtx)
# make the spending
scriptsig = CScript([cds_signature])
tx = CTransaction()
tx.vin.append(CTxIn(COutPoint(fundtx.sha256, 1), scriptsig))
input_index = 2
remaining_sigcheck = sigcheckcount
while remaining_sigcheck > 30:
tx.vin.append(
CTxIn(COutPoint(fundtx.sha256, input_index), scriptsig))
remaining_sigcheck -= 30
input_index += 1
tx.vout.append(CTxOut(0, CScript([OP_RETURN])))
pad_tx(tx)
tx.rehash()
return tx
# Create transactions with many sigchecks.
good_tx = make_spend(MAX_TX_SIGCHECK)
bad_tx = make_spend(MAX_TX_SIGCHECK + 1)
tip = self.build_block(tip, fundings)
node.p2p.send_blocks_and_test([tip], node)
# Both tx are accepted before the activation.
pre_activation_sigcheck_block = self.build_block(
tip, [good_tx, bad_tx])
node.p2p.send_blocks_and_test([pre_activation_sigcheck_block], node)
node.invalidateblock(pre_activation_sigcheck_block.hash)
# Activation tests
self.log.info("Approach to just before upgrade activation")
# Move our clock to the uprade time so we will accept such
# future-timestamped blocks.
node.setmocktime(SIGCHECKS_ACTIVATION_TIME + 10)
# Mine six blocks with timestamp starting at
# SIGCHECKS_ACTIVATION_TIME-1
blocks = []
for i in range(-1, 5):
tip = self.build_block(tip, nTime=SIGCHECKS_ACTIVATION_TIME + i)
blocks.append(tip)
node.p2p.send_blocks_and_test(blocks, node)
assert_equal(node.getblockchaininfo()['mediantime'],
SIGCHECKS_ACTIVATION_TIME - 1)
self.log.info(
"The next block will activate, but the activation block itself must follow old rules"
)
# Send the 50 txes and get the node to mine as many as possible (it should do all)
# The node is happy mining and validating a 10000 sigcheck block before
# activation.
node.p2p.send_txs_and_test(submittxes_1, node)
[blockhash
] = node.generatetoaddress(1,
node.get_deterministic_priv_key().address)
assert_equal(set(node.getblock(blockhash, 1)["tx"][1:]),
{t.hash
for t in submittxes_1})
# We have activated, but let's invalidate that.
assert_equal(node.getblockchaininfo()['mediantime'],
SIGCHECKS_ACTIVATION_TIME)
node.invalidateblock(blockhash)
# Try again manually and invalidate that too
goodblock = self.build_block(tip, submittxes_1)
node.p2p.send_blocks_and_test([goodblock], node)
node.invalidateblock(goodblock.hash)
# All transactions should be back in mempool.
assert_equal(set(node.getrawmempool()), {t.hash for t in submittxes_1})
self.log.info("Mine the activation block itself")
tip = self.build_block(tip)
node.p2p.send_blocks_and_test([tip], node)
self.log.info("We have activated!")
assert_equal(node.getblockchaininfo()['mediantime'],
SIGCHECKS_ACTIVATION_TIME)
self.log.info(
"Try a block with a transaction going over the limit (limit: {})".
format(MAX_TX_SIGCHECK))
bad_tx_block = self.build_block(tip, [bad_tx])
check_for_ban_on_rejected_block(
node, bad_tx_block, reject_reason=BLOCK_SIGCHECKS_PARALLEL_ERROR)
self.log.info(
"Try a block with a transaction just under the limit (limit: {})".
format(MAX_TX_SIGCHECK))
good_tx_block = self.build_block(tip, [good_tx])
node.p2p.send_blocks_and_test([good_tx_block], node)
node.invalidateblock(good_tx_block.hash)
# save this tip for later
# ~ upgrade_block = tip
# Transactions still in pool:
assert_equal(set(node.getrawmempool()), {t.hash for t in submittxes_1})
self.log.info(
"Try sending 10000-sigcheck blocks after activation (limit: {})".
format(MAXBLOCKSIZE // BLOCK_MAXBYTES_MAXSIGCHECKS_RATIO))
# Send block with same txes we just tried before activation
badblock = self.build_block(tip, submittxes_1)
check_for_ban_on_rejected_block(
node, badblock, reject_reason=BLOCK_SIGCHECKS_CACHED_ERROR)
self.log.info(
"There are too many sigchecks in mempool to mine in a single block. Make sure the node won't mine invalid blocks."
)
node.generatetoaddress(1, node.get_deterministic_priv_key().address)
tip = self.getbestblock(node)
# only 39 txes got mined.
assert_equal(len(node.getrawmempool()), 11)
self.log.info(
"Try sending 10000-sigcheck block with fresh transactions after activation (limit: {})"
.format(MAXBLOCKSIZE // BLOCK_MAXBYTES_MAXSIGCHECKS_RATIO))
# Note: in the following tests we'll be bumping timestamp in order
# to bypass any kind of 'bad block' cache on the node, and get a
# fresh evaluation each time.
# Try another block with 10000 sigchecks but all fresh transactions
badblock = self.build_block(tip,
submittxes_2,
nTime=SIGCHECKS_ACTIVATION_TIME + 5)
check_for_ban_on_rejected_block(
node, badblock, reject_reason=BLOCK_SIGCHECKS_PARALLEL_ERROR)
# Send the same txes again with different block hash. Currently we don't
# cache valid transactions in invalid blocks so nothing changes.
badblock = self.build_block(tip,
submittxes_2,
nTime=SIGCHECKS_ACTIVATION_TIME + 6)
check_for_ban_on_rejected_block(
node, badblock, reject_reason=BLOCK_SIGCHECKS_PARALLEL_ERROR)
# Put all the txes in mempool, in order to get them cached:
node.p2p.send_txs_and_test(submittxes_2, node)
# Send them again, the node still doesn't like it. But the log
# error message has now changed because the txes failed from cache.
badblock = self.build_block(tip,
submittxes_2,
nTime=SIGCHECKS_ACTIVATION_TIME + 7)
check_for_ban_on_rejected_block(
node, badblock, reject_reason=BLOCK_SIGCHECKS_CACHED_ERROR)
self.log.info(
"Try sending 8000-sigcheck block after activation (limit: {})".
format(MAXBLOCKSIZE // BLOCK_MAXBYTES_MAXSIGCHECKS_RATIO))
# redundant, but just to mirror the following test...
node.setexcessiveblock(MAXBLOCKSIZE)
badblock = self.build_block(tip,
submittxes_2[:40],
nTime=SIGCHECKS_ACTIVATION_TIME + 5)
check_for_ban_on_rejected_block(
node, badblock, reject_reason=BLOCK_SIGCHECKS_CACHED_ERROR)
self.log.info(
"Bump the excessiveblocksize limit by 1 byte, and send another block with same txes (new sigchecks limit: {})"
.format((MAXBLOCKSIZE + 1) // BLOCK_MAXBYTES_MAXSIGCHECKS_RATIO))
node.setexcessiveblock(MAXBLOCKSIZE + 1)
tip = self.build_block(tip,
submittxes_2[:40],
nTime=SIGCHECKS_ACTIVATION_TIME + 6)
# It should succeed now since limit should be 8000.
node.p2p.send_blocks_and_test([tip], node)
def test_sequence_lock_unconfirmed_inputs(self):
# Store height so we can easily reset the chain at the end of the test
cur_height = self.nodes[0].getblockcount()
# Create a mempool tx.
txid = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), 2)
tx1 = FromHex(CTransaction(), self.nodes[0].getrawtransaction(txid))
tx1.rehash()
# Anyone-can-spend mempool tx.
# Sequence lock of 0 should pass.
tx2 = CTransaction()
tx2.nVersion = 2
tx2.vin = [CTxIn(COutPoint(tx1.sha256, 0), nSequence=0)]
tx2.vout = [CTxOut(int(tx1.vout[0].nValue - self.relayfee*COIN), CScript([b'a']))]
tx2_raw = self.nodes[0].signrawtransactionwithwallet(ToHex(tx2))["hex"]
tx2 = FromHex(tx2, tx2_raw)
tx2.rehash()
self.nodes[0].sendrawtransaction(tx2_raw)
# Create a spend of the 0th output of orig_tx with a sequence lock
# of 1, and test what happens when submitting.
# orig_tx.vout[0] must be an anyone-can-spend output
def test_nonzero_locks(orig_tx, node, relayfee, use_height_lock):
sequence_value = 1
if not use_height_lock:
sequence_value |= SEQUENCE_LOCKTIME_TYPE_FLAG
tx = CTransaction()
tx.nVersion = 2
tx.vin = [CTxIn(COutPoint(orig_tx.sha256, 0), nSequence=sequence_value)]
tx.vout = [CTxOut(int(orig_tx.vout[0].nValue - relayfee * COIN), CScript([b'a' * 35]))]
tx.rehash()
if (orig_tx.hash in node.getrawmempool()):
# sendrawtransaction should fail if the tx is in the mempool
assert_raises_rpc_error(-26, NOT_FINAL_ERROR, node.sendrawtransaction, ToHex(tx))
else:
# sendrawtransaction should succeed if the tx is not in the mempool
node.sendrawtransaction(ToHex(tx))
return tx
test_nonzero_locks(tx2, self.nodes[0], self.relayfee, use_height_lock=True)
test_nonzero_locks(tx2, self.nodes[0], self.relayfee, use_height_lock=False)
# Now mine some blocks, but make sure tx2 doesn't get mined.
# Use prioritisetransaction to lower the effective feerate to 0
self.nodes[0].prioritisetransaction(txid=tx2.hash, fee_delta=int(-self.relayfee*COIN))
cur_time = int(time.time())
for i in range(10):
self.nodes[0].setmocktime(cur_time + 600)
self.nodes[0].generate(1)
cur_time += 600
assert tx2.hash in self.nodes[0].getrawmempool()
test_nonzero_locks(tx2, self.nodes[0], self.relayfee, use_height_lock=True)
test_nonzero_locks(tx2, self.nodes[0], self.relayfee, use_height_lock=False)
# Mine tx2, and then try again
self.nodes[0].prioritisetransaction(txid=tx2.hash, fee_delta=int(self.relayfee*COIN))
# Advance the time on the node so that we can test timelocks
self.nodes[0].setmocktime(cur_time+600)
self.nodes[0].generate(1)
assert tx2.hash not in self.nodes[0].getrawmempool()
# Now that tx2 is not in the mempool, a sequence locked spend should
# succeed
tx3 = test_nonzero_locks(tx2, self.nodes[0], self.relayfee, use_height_lock=False)
assert tx3.hash in self.nodes[0].getrawmempool()
self.nodes[0].generate(1)
assert tx3.hash not in self.nodes[0].getrawmempool()
# One more test, this time using height locks
tx4 = test_nonzero_locks(tx3, self.nodes[0], self.relayfee, use_height_lock=True)
assert tx4.hash in self.nodes[0].getrawmempool()
# Now try combining confirmed and unconfirmed inputs
tx5 = test_nonzero_locks(tx4, self.nodes[0], self.relayfee, use_height_lock=True)
assert tx5.hash not in self.nodes[0].getrawmempool()
utxos = self.nodes[0].listunspent()
tx5.vin.append(CTxIn(COutPoint(int(utxos[0]["txid"], 16), utxos[0]["vout"]), nSequence=1))
tx5.vout[0].nValue += int(utxos[0]["amount"]*COIN)
raw_tx5 = self.nodes[0].signrawtransactionwithwallet(ToHex(tx5))["hex"]
assert_raises_rpc_error(-26, NOT_FINAL_ERROR, self.nodes[0].sendrawtransaction, raw_tx5)
# Test mempool-BIP68 consistency after reorg
#
# State of the transactions in the last blocks:
# ... -> [ tx2 ] -> [ tx3 ]
# tip-1 tip
# And currently tx4 is in the mempool.
#
# If we invalidate the tip, tx3 should get added to the mempool, causing
# tx4 to be removed (fails sequence-lock).
self.nodes[0].invalidateblock(self.nodes[0].getbestblockhash())
assert tx4.hash not in self.nodes[0].getrawmempool()
assert tx3.hash in self.nodes[0].getrawmempool()
# Now mine 2 empty blocks to reorg out the current tip (labeled tip-1 in
# diagram above).
# This would cause tx2 to be added back to the mempool, which in turn causes
# tx3 to be removed.
tip = int(self.nodes[0].getblockhash(self.nodes[0].getblockcount()-1), 16)
height = self.nodes[0].getblockcount()
for i in range(2):
block = create_block(tip, create_coinbase(height), cur_time)
block.set_base_version(3)
block.rehash()
block.solve()
tip = block.sha256
height += 1
self.nodes[0].submitblock(ToHex(block))
cur_time += 1
mempool = self.nodes[0].getrawmempool()
assert tx3.hash not in mempool
assert tx2.hash in mempool
# Reset the chain and get rid of the mocktimed-blocks
self.nodes[0].setmocktime(0)
self.nodes[0].invalidateblock(self.nodes[0].getblockhash(cur_height+1))
self.nodes[0].generate(10)
def run_test(self):
node = self.nodes[0]
self.log.info('Start with empty mempool, and 200 blocks')
self.mempool_size = 0
assert_equal(node.getblockcount(), 200)
assert_equal(node.getmempoolinfo()['size'], self.mempool_size)
coins = node.listunspent()
self.log.info('Should not accept garbage to testmempoolaccept')
assert_raises_rpc_error(-3, 'Expected type array, got string', lambda: node.testmempoolaccept(rawtxs='ff00baar'))
assert_raises_rpc_error(-8, 'Array must contain exactly one raw transaction for now', lambda: node.testmempoolaccept(rawtxs=['ff00baar', 'ff22']))
assert_raises_rpc_error(-22, 'TX decode failed', lambda: node.testmempoolaccept(rawtxs=['ff00baar']))
self.log.info('A transaction already in the blockchain')
coin = coins.pop() # Pick a random coin(base) to spend
raw_tx_in_block = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{'txid': coin['txid'], 'vout': coin['vout']}],
outputs=[{node.getnewaddress(): 0.3}, {node.getnewaddress(): 49}],
))['hex']
txid_in_block = node.sendrawtransaction(hexstring=raw_tx_in_block, maxfeerate=0)
node.generate(1)
self.mempool_size = 0
self.check_mempool_result(
result_expected=[{'txid': txid_in_block, 'allowed': False, 'reject-reason': 'txn-already-known'}],
rawtxs=[raw_tx_in_block],
)
self.log.info('A transaction not in the mempool')
fee = Decimal('0.000007')
raw_tx_0 = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{"txid": txid_in_block, "vout": 0, "sequence": BIP125_SEQUENCE_NUMBER}], # RBF is used later
outputs=[{node.getnewaddress(): Decimal('0.3') - fee}],
))['hex']
tx = CTransaction()
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
txid_0 = tx.rehash()
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': True, 'vsize': tx.get_vsize(), 'fees': {'base': fee}}],
rawtxs=[raw_tx_0],
)
self.log.info('A final transaction not in the mempool')
coin = coins.pop() # Pick a random coin(base) to spend
output_amount = Decimal('0.025')
raw_tx_final = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{'txid': coin['txid'], 'vout': coin['vout'], "sequence": 0xffffffff}], # SEQUENCE_FINAL
outputs=[{node.getnewaddress(): output_amount}],
locktime=node.getblockcount() + 2000, # Can be anything
))['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_final)))
fee_expected = coin['amount'] - output_amount
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': True, 'vsize': tx.get_vsize(), 'fees': {'base': fee_expected}}],
rawtxs=[tx.serialize().hex()],
maxfeerate=0,
)
node.sendrawtransaction(hexstring=raw_tx_final, maxfeerate=0)
self.mempool_size += 1
self.log.info('A transaction in the mempool')
node.sendrawtransaction(hexstring=raw_tx_0)
self.mempool_size += 1
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': False, 'reject-reason': 'txn-already-in-mempool'}],
rawtxs=[raw_tx_0],
)
self.log.info('A transaction that replaces a mempool transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vout[0].nValue -= int(fee * COIN) # Double the fee
tx.vin[0].nSequence = BIP125_SEQUENCE_NUMBER + 1 # Now, opt out of RBF
raw_tx_0 = node.signrawtransactionwithwallet(tx.serialize().hex())['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
txid_0 = tx.rehash()
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': True, 'vsize': tx.get_vsize(), 'fees': {'base': (2 * fee)}}],
rawtxs=[raw_tx_0],
)
self.log.info('A transaction that conflicts with an unconfirmed tx')
# Send the transaction that replaces the mempool transaction and opts out of replaceability
node.sendrawtransaction(hexstring=tx.serialize().hex(), maxfeerate=0)
# take original raw_tx_0
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vout[0].nValue -= int(4 * fee * COIN) # Set more fee
# skip re-signing the tx
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'txn-mempool-conflict'}],
rawtxs=[tx.serialize().hex()],
maxfeerate=0,
)
self.log.info('A transaction with missing inputs, that never existed')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vin[0].prevout = COutPoint(hash=int('ff' * 32, 16), n=14)
# skip re-signing the tx
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'missing-inputs'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with missing inputs, that existed once in the past')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vin[0].prevout.n = 1 # Set vout to 1, to spend the other outpoint (49 coins) of the in-chain-tx we want to double spend
raw_tx_1 = node.signrawtransactionwithwallet(tx.serialize().hex())['hex']
txid_1 = node.sendrawtransaction(hexstring=raw_tx_1, maxfeerate=0)
# Now spend both to "clearly hide" the outputs, ie. remove the coins from the utxo set by spending them
raw_tx_spend_both = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[
{'txid': txid_0, 'vout': 0},
{'txid': txid_1, 'vout': 0},
],
outputs=[{node.getnewaddress(): 0.1}]
))['hex']
txid_spend_both = node.sendrawtransaction(hexstring=raw_tx_spend_both, maxfeerate=0)
node.generate(1)
self.mempool_size = 0
# Now see if we can add the coins back to the utxo set by sending the exact txs again
self.check_mempool_result(
result_expected=[{'txid': txid_0, 'allowed': False, 'reject-reason': 'missing-inputs'}],
rawtxs=[raw_tx_0],
)
self.check_mempool_result(
result_expected=[{'txid': txid_1, 'allowed': False, 'reject-reason': 'missing-inputs'}],
rawtxs=[raw_tx_1],
)
self.log.info('Create a signed "reference" tx for later use')
raw_tx_reference = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{'txid': txid_spend_both, 'vout': 0}],
outputs=[{node.getnewaddress(): 0.05}],
))['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
# Reference tx should be valid on itself
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': True, 'vsize': tx.get_vsize(), 'fees': { 'base': Decimal('0.1') - Decimal('0.05')}}],
rawtxs=[tx.serialize().hex()],
maxfeerate=0,
)
self.log.info('A transaction with no outputs')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout = []
# Skip re-signing the transaction for context independent checks from now on
# tx.deserialize(BytesIO(hex_str_to_bytes(node.signrawtransactionwithwallet(tx.serialize().hex())['hex'])))
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-txns-vout-empty'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A really large transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin = [tx.vin[0]] * math.ceil(MAX_BLOCK_BASE_SIZE / len(tx.vin[0].serialize()))
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-txns-oversize'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with negative output value')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].nValue *= -1
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-txns-vout-negative'}],
rawtxs=[tx.serialize().hex()],
)
# The following two validations prevent overflow of the output amounts (see CVE-2010-5139).
self.log.info('A transaction with too large output value')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].nValue = MAX_MONEY + 1
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-txns-vout-toolarge'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with too large sum of output values')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout = [tx.vout[0]] * 2
tx.vout[0].nValue = MAX_MONEY
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-txns-txouttotal-toolarge'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A transaction with duplicate inputs')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin = [tx.vin[0]] * 2
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-txns-inputs-duplicate'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('A coinbase transaction')
# Pick the input of the first tx we signed, so it has to be a coinbase tx
raw_tx_coinbase_spent = node.getrawtransaction(txid=node.decoderawtransaction(hexstring=raw_tx_in_block)['vin'][0]['txid'])
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_coinbase_spent)))
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'coinbase'}],
rawtxs=[tx.serialize().hex()],
)
self.log.info('Some nonstandard transactions')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.nVersion = 3 # A version currently non-standard
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'version'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].scriptPubKey = CScript([OP_0]) # Some non-standard script
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'scriptpubkey'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
key = ECKey()
key.generate()
pubkey = key.get_pubkey().get_bytes()
tx.vout[0].scriptPubKey = CScript([OP_2, pubkey, pubkey, pubkey, OP_3, OP_CHECKMULTISIG]) # Some bare multisig script (2-of-3)
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bare-multisig'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].scriptSig = CScript([OP_HASH160]) # Some not-pushonly scriptSig
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'scriptsig-not-pushonly'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].scriptSig = CScript([b'a' * 1648]) # Some too large scriptSig (>1650 bytes)
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'scriptsig-size'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
output_p2sh_burn = CTxOut(nValue=540, scriptPubKey=CScript([OP_HASH160, hash160(b'burn'), OP_EQUAL]))
num_scripts = 10000 // len(output_p2sh_burn.serialize()) # Use enough outputs to make the tx too large for our policy
tx.vout = [output_p2sh_burn] * num_scripts
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'tx-size'}],
rawtxs=[tx.serialize().hex()],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0] = output_p2sh_burn
tx.vout[0].nValue -= 1 # Make output smaller, such that it is dust for our policy
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'dust'}],
rawtxs=[tx.serialize().hex()],
)
# Unlike upstream, Xaya allows multiple OP_RETURN outputs. So no test for this.
self.log.info('A timelocked transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].nSequence -= 1 # Should be non-max, so locktime is not ignored
tx.nLockTime = node.getblockcount() + 1
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'non-final'}],
rawtxs=[tx.serialize().hex()],
)
# FIXME: Enable once Namecoin has BIP68 enabled.
return
self.log.info('A transaction that is locked by BIP68 sequence logic')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].nSequence = 2 # We could include it in the second block mined from now, but not the very next one
# Can skip re-signing the tx because of early rejection
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'non-BIP68-final'}],
rawtxs=[tx.serialize().hex()],
maxfeerate=0,
)
def test_too_many_replacements(self):
"""Replacements that evict too many transactions are rejected"""
# Try directly replacing more than MAX_REPLACEMENT_LIMIT
# transactions
# Start by creating a single transaction with many outputs
initial_nValue = 10 * COIN
utxo = make_utxo(self.nodes[0], initial_nValue)
fee = int(0.0001 * COIN)
split_value = int((initial_nValue - fee) / (MAX_REPLACEMENT_LIMIT + 1))
outputs = []
for i in range(MAX_REPLACEMENT_LIMIT + 1):
outputs.append(CTxOut(split_value, CScript([1])))
splitting_tx = CTransaction()
splitting_tx.vin = [CTxIn(utxo, nSequence=0)]
splitting_tx.vout = outputs + [
CTxOut(
int(initial_nValue -
(MAX_REPLACEMENT_LIMIT + 1) * split_value))
]
splitting_tx_hex = txToHex(splitting_tx)
txid = self.nodes[0].sendrawtransaction(splitting_tx_hex, True)
txid = int(txid, 16)
# Now spend each of those outputs individually
for i in range(MAX_REPLACEMENT_LIMIT + 1):
tx_i = CTransaction()
tx_i.vin = [CTxIn(COutPoint(txid, i), nSequence=0)]
tx_i.vout = [
CTxOut(split_value - fee, CScript([b'a' * 35])),
CTxOut(fee)
]
tx_i_hex = txToHex(tx_i)
self.nodes[0].sendrawtransaction(tx_i_hex, True)
# Now create doublespend of the whole lot; should fail.
# Need a big enough fee to cover all spending transactions and have
# a higher fee rate
double_spend_value = (split_value -
100 * fee) * (MAX_REPLACEMENT_LIMIT + 1)
inputs = []
for i in range(MAX_REPLACEMENT_LIMIT + 1):
inputs.append(CTxIn(COutPoint(txid, i), nSequence=0))
double_tx = CTransaction()
double_tx.vin = inputs
double_tx.vout = [
CTxOut(double_spend_value, CScript([b'a'])),
CTxOut(
int(split_value * (MAX_REPLACEMENT_LIMIT + 1) -
double_spend_value))
]
double_tx_hex = txToHex(double_tx)
# This will raise an exception
assert_raises_rpc_error(-26, "too many potential replacements",
self.nodes[0].sendrawtransaction,
double_tx_hex, True)
# If we remove an input, it should pass
double_tx = CTransaction()
double_tx.vin = inputs[0:-1]
double_tx.vout = [
CTxOut(double_spend_value, CScript([b'a'])),
CTxOut(
int(split_value * (MAX_REPLACEMENT_LIMIT) -
double_spend_value))
]
double_tx_hex = txToHex(double_tx)
self.nodes[0].sendrawtransaction(double_tx_hex, True)
def test_prioritised_transactions(self):
# Ensure that fee deltas used via prioritisetransaction are
# correctly used by replacement logic
# 1. Check that feeperkb uses modified fees
tx0_outpoint = make_utxo(self.nodes[0], int(1.1 * COIN))
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
# Higher fee, but the actual fee per KB is much lower.
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.001 * COIN), CScript([b'a' * 740000]))]
tx1b_hex = txToHex(tx1b)
# Verify tx1b cannot replace tx1a.
assert_raises_rpc_error(-26, "insufficient fee",
self.nodes[0].sendrawtransaction, tx1b_hex,
True)
# Use prioritisetransaction to set tx1a's fee to 0.
self.nodes[0].prioritisetransaction(txid=tx1a_txid,
fee_delta=int(-0.1 * COIN))
# Now tx1b should be able to replace tx1a
tx1b_txid = self.nodes[0].sendrawtransaction(tx1b_hex, True)
assert (tx1b_txid in self.nodes[0].getrawmempool())
# 2. Check that absolute fee checks use modified fee.
tx1_outpoint = make_utxo(self.nodes[0], int(1.1 * COIN))
tx2a = CTransaction()
tx2a.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx2a_hex = txToHex(tx2a)
self.nodes[0].sendrawtransaction(tx2a_hex, True)
# Lower fee, but we'll prioritise it
tx2b = CTransaction()
tx2b.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2b.vout = [CTxOut(int(1.01 * COIN), CScript([b'a' * 35]))]
tx2b.rehash()
tx2b_hex = txToHex(tx2b)
# Verify tx2b cannot replace tx2a.
assert_raises_rpc_error(-26, "insufficient fee",
self.nodes[0].sendrawtransaction, tx2b_hex,
True)
# Now prioritise tx2b to have a higher modified fee
self.nodes[0].prioritisetransaction(txid=tx2b.hash,
fee_delta=int(0.1 * COIN))
# tx2b should now be accepted
tx2b_txid = self.nodes[0].sendrawtransaction(tx2b_hex, True)
assert (tx2b_txid in self.nodes[0].getrawmempool())
def test_doublespend_tree(self):
"""Doublespend of a big tree of transactions"""
initial_nValue = 50 * COIN
tx0_outpoint = make_utxo(self.nodes[0], initial_nValue)
def branch(prevout,
initial_value,
max_txs,
tree_width=5,
fee=0.0001 * COIN,
_total_txs=None):
if _total_txs is None:
_total_txs = [0]
if _total_txs[0] >= max_txs:
return
txout_value = (initial_value - fee) // tree_width
if txout_value < fee:
return
vout = [
CTxOut(txout_value, CScript([i + 1]))
for i in range(tree_width)
]
tx = CTransaction()
tx.vin = [CTxIn(prevout, nSequence=0)]
tx.vout = vout
tx_hex = txToHex(tx)
assert len(tx.serialize()) < 100000
txid = self.nodes[0].sendrawtransaction(tx_hex, 0)
yield tx
_total_txs[0] += 1
txid = int(txid, 16)
for i, txout in enumerate(tx.vout):
for x in branch(COutPoint(txid, i),
txout_value,
max_txs,
tree_width=tree_width,
fee=fee,
_total_txs=_total_txs):
yield x
fee = int(0.0001 * COIN)
n = MAX_REPLACEMENT_LIMIT
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)
# Attempt double-spend, will fail because too little fee paid
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [CTxOut(initial_nValue - fee * n, DUMMY_P2WPKH_SCRIPT)]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee",
self.nodes[0].sendrawtransaction, dbl_tx_hex,
0)
# 1 BTCI fee is enough
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [
CTxOut(initial_nValue - fee * n - 1 * COIN, DUMMY_P2WPKH_SCRIPT)
]
dbl_tx_hex = txToHex(dbl_tx)
self.nodes[0].sendrawtransaction(dbl_tx_hex, 0)
mempool = self.nodes[0].getrawmempool()
for tx in tree_txs:
tx.rehash()
assert tx.hash not in mempool
# Try again, but with more total transactions than the "max txs
# double-spent at once" anti-DoS limit.
for n in (MAX_REPLACEMENT_LIMIT + 1, MAX_REPLACEMENT_LIMIT * 2):
fee = int(0.0001 * COIN)
tx0_outpoint = make_utxo(self.nodes[0], initial_nValue)
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
dbl_tx.vout = [
CTxOut(initial_nValue - 2 * fee * n, DUMMY_P2WPKH_SCRIPT)
]
dbl_tx_hex = txToHex(dbl_tx)
# This will raise an exception
assert_raises_rpc_error(-26, "too many potential replacements",
self.nodes[0].sendrawtransaction,
dbl_tx_hex, 0)
for tx in tree_txs:
tx.rehash()
self.nodes[0].getrawtransaction(tx.hash)
def run_test(self):
node = self.nodes[0]
self.log.info('Start with empty mempool, and 200 blocks')
self.mempool_size = 0
wait_until(lambda: node.getblockcount() == 200)
assert_equal(node.getmempoolinfo()['size'], self.mempool_size)
self.log.info('Should not accept garbage to testmempoolaccept')
assert_raises_rpc_error(
-3, 'Expected type array, got string',
lambda: node.testmempoolaccept(rawtxs='ff00baar'))
assert_raises_rpc_error(
-8, 'Array must contain exactly one raw transaction for now',
lambda: node.testmempoolaccept(rawtxs=['ff00baar', 'ff22']))
assert_raises_rpc_error(
-22, 'TX decode failed',
lambda: node.testmempoolaccept(rawtxs=['ff00baar']))
self.log.info('A transaction already in the blockchain')
coin = node.listunspent()[0] # Pick a random coin(base) to spend
raw_tx_in_block = node.signrawtransactionwithwallet(
node.createrawtransaction(
inputs=[{
'txid': coin['txid'],
'vout': coin['vout']
}],
outputs=[{
node.getnewaddress(): 0.3
}, {
node.getnewaddress(): 49
}],
))['hex']
txid_in_block = node.sendrawtransaction(hexstring=raw_tx_in_block,
allowhighfees=True)
node.generate(1)
self.check_mempool_result(
result_expected=[{
'txid': txid_in_block,
'allowed': False,
'reject-reason': '18: txn-already-known'
}],
rawtxs=[raw_tx_in_block],
)
self.log.info('A transaction not in the mempool')
fee = 0.00000700
raw_tx_0 = node.signrawtransactionwithwallet(
node.createrawtransaction(
inputs=[{
"txid": txid_in_block,
"vout": 0,
"sequence": BIP125_SEQUENCE_NUMBER
}], # RBF is used later
outputs=[{
node.getnewaddress(): 0.3 - fee
}],
))['hex']
tx = CTransaction()
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
txid_0 = tx.rehash()
self.check_mempool_result(
result_expected=[{
'txid': txid_0,
'allowed': True
}],
rawtxs=[raw_tx_0],
)
self.log.info('A transaction in the mempool')
node.sendrawtransaction(hexstring=raw_tx_0)
self.mempool_size = 1
self.check_mempool_result(
result_expected=[{
'txid': txid_0,
'allowed': False,
'reject-reason': '18: txn-already-in-mempool'
}],
rawtxs=[raw_tx_0],
)
self.log.info('A transaction that replaces a mempool transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vout[0].nValue -= int(fee * COIN) # Double the fee
tx.vin[0].nSequence = BIP125_SEQUENCE_NUMBER + 1 # Now, opt out of RBF
raw_tx_0 = node.signrawtransactionwithwallet(
bytes_to_hex_str(tx.serialize()))['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
txid_0 = tx.rehash()
self.check_mempool_result(
result_expected=[{
'txid': txid_0,
'allowed': True
}],
rawtxs=[raw_tx_0],
)
self.log.info('A transaction that conflicts with an unconfirmed tx')
# Send the transaction that replaces the mempool transaction and opts out of replaceability
node.sendrawtransaction(hexstring=bytes_to_hex_str(tx.serialize()),
allowhighfees=True)
# take original raw_tx_0
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vout[0].nValue -= int(4 * fee * COIN) # Set more fee
# skip re-signing the tx
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '18: txn-mempool-conflict'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
allowhighfees=True,
)
self.log.info('A transaction with missing inputs, that never existed')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vin[0].prevout = COutPoint(hash=int('ff' * 32, 16), n=14)
# skip re-signing the tx
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': 'missing-inputs'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info(
'A transaction with missing inputs, that existed once in the past')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_0)))
tx.vin[
0].prevout.n = 1 # Set vout to 1, to spend the other outpoint (49 coins) of the in-chain-tx we want to double spend
raw_tx_1 = node.signrawtransactionwithwallet(
bytes_to_hex_str(tx.serialize()))['hex']
txid_1 = node.sendrawtransaction(hexstring=raw_tx_1,
allowhighfees=True)
# Now spend both to "clearly hide" the outputs, ie. remove the coins from the utxo set by spending them
raw_tx_spend_both = node.signrawtransactionwithwallet(
node.createrawtransaction(inputs=[
{
'txid': txid_0,
'vout': 0
},
{
'txid': txid_1,
'vout': 0
},
],
outputs=[{
node.getnewaddress(): 0.1
}]))['hex']
txid_spend_both = node.sendrawtransaction(hexstring=raw_tx_spend_both,
allowhighfees=True)
node.generate(1)
self.mempool_size = 0
# Now see if we can add the coins back to the utxo set by sending the exact txs again
self.check_mempool_result(
result_expected=[{
'txid': txid_0,
'allowed': False,
'reject-reason': 'missing-inputs'
}],
rawtxs=[raw_tx_0],
)
self.check_mempool_result(
result_expected=[{
'txid': txid_1,
'allowed': False,
'reject-reason': 'missing-inputs'
}],
rawtxs=[raw_tx_1],
)
self.log.info('Create a signed "reference" tx for later use')
raw_tx_reference = node.signrawtransactionwithwallet(
node.createrawtransaction(
inputs=[{
'txid': txid_spend_both,
'vout': 0
}],
outputs=[{
node.getnewaddress(): 0.05
}],
))['hex']
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
# Reference tx should be valid on itself
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': True
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A transaction with no outputs')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout = []
# Skip re-signing the transaction for context independent checks from now on
# tx.deserialize(BytesIO(hex_str_to_bytes(node.signrawtransactionwithwallet(bytes_to_hex_str(tx.serialize()))['hex'])))
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '16: bad-txns-vout-empty'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A really large transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin = [tx.vin[0]
] * (MAX_BLOCK_BASE_SIZE // len(tx.vin[0].serialize()))
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '16: bad-txns-oversize'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A transaction with negative output value')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].nValue *= -1
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '16: bad-txns-vout-negative'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A transaction with too large output value')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].nValue = 21000000 * COIN + 1
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '16: bad-txns-vout-toolarge'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A transaction with too large sum of output values')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout = [tx.vout[0]] * 2
tx.vout[0].nValue = 21000000 * COIN
self.check_mempool_result(
result_expected=[{
'txid':
tx.rehash(),
'allowed':
False,
'reject-reason':
'16: bad-txns-txouttotal-toolarge'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A transaction with duplicate inputs')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin = [tx.vin[0]] * 2
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '16: bad-txns-inputs-duplicate'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A coinbase transaction')
# Pick the input of the first tx we signed, so it has to be a coinbase tx
raw_tx_coinbase_spent = node.getrawtransaction(
txid=node.decoderawtransaction(
hexstring=raw_tx_in_block)['vin'][0]['txid'])
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_coinbase_spent)))
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '16: coinbase'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('Some nonstandard transactions')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.nVersion = 3 # A version currently non-standard
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: version'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].scriptPubKey = CScript([OP_0]) # Some non-standard script
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: scriptpubkey'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[0].scriptSig = CScript([OP_HASH160
]) # Some not-pushonly scriptSig
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: scriptsig-not-pushonly'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
output_p2sh_burn = CTxOut(nValue=540,
scriptPubKey=CScript(
[OP_HASH160,
hash160(b'burn'), OP_EQUAL]))
num_scripts = 100000 // len(output_p2sh_burn.serialize(
)) # Use enough outputs to make the tx too large for our policy
tx.vout = [output_p2sh_burn] * num_scripts
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: tx-size'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0] = output_p2sh_burn
tx.vout[
0].nValue -= 1 # Make output smaller, such that it is dust for our policy
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: dust'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vout[0].scriptPubKey = CScript([OP_RETURN, b'\xff'])
tx.vout = [tx.vout[0]] * 2
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: multi-op-return'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A timelocked transaction')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[
0].nSequence -= 1 # Should be non-max, so locktime is not ignored
tx.nLockTime = node.getblockcount() + 1
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: non-final'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
)
self.log.info('A transaction that is locked by BIP68 sequence logic')
tx.deserialize(BytesIO(hex_str_to_bytes(raw_tx_reference)))
tx.vin[
0].nSequence = 2 # We could include it in the second block mined from now, but not the very next one
# Can skip re-signing the tx because of early rejection
self.check_mempool_result(
result_expected=[{
'txid': tx.rehash(),
'allowed': False,
'reject-reason': '64: non-BIP68-final'
}],
rawtxs=[bytes_to_hex_str(tx.serialize())],
allowhighfees=True,
)
def test_opt_in(self):
"""Replacing should only work if orig tx opted in"""
tx0_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
# Create a non-opting in transaction
tx1a = CTransaction()
tx1a.vin = [CTxIn(tx0_outpoint, nSequence=0xffffffff)]
tx1a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx1a_hex = txToHex(tx1a)
tx1a_txid = self.nodes[0].sendrawtransaction(tx1a_hex, True)
# This transaction isn't shown as replaceable
assert_equal(self.nodes[0].getmempoolentry(tx1a_txid)['bip125-replaceable'], False)
# Shouldn't be able to double-spend
tx1b = CTransaction()
tx1b.vin = [CTxIn(tx0_outpoint, nSequence=0)]
tx1b.vout = [CTxOut(int(0.9 * COIN), CScript([b'b' * 35]))]
tx1b_hex = txToHex(tx1b)
# This will raise an exception
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[0].sendrawtransaction, tx1b_hex, True)
tx1_outpoint = make_utxo(self.nodes[0], int(1.1*COIN))
# Create a different non-opting in transaction
tx2a = CTransaction()
tx2a.vin = [CTxIn(tx1_outpoint, nSequence=0xfffffffe)]
tx2a.vout = [CTxOut(1 * COIN, CScript([b'a' * 35]))]
tx2a_hex = txToHex(tx2a)
tx2a_txid = self.nodes[0].sendrawtransaction(tx2a_hex, True)
# Still shouldn't be able to double-spend
tx2b = CTransaction()
tx2b.vin = [CTxIn(tx1_outpoint, nSequence=0)]
tx2b.vout = [CTxOut(int(0.9 * COIN), CScript([b'b' * 35]))]
tx2b_hex = txToHex(tx2b)
# This will raise an exception
assert_raises_rpc_error(-26, "txn-mempool-conflict", self.nodes[0].sendrawtransaction, tx2b_hex, True)
# Now create a new transaction that spends from tx1a and tx2a
# opt-in on one of the inputs
# Transaction should be replaceable on either input
tx1a_txid = int(tx1a_txid, 16)
tx2a_txid = int(tx2a_txid, 16)
tx3a = CTransaction()
tx3a.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0xffffffff),
CTxIn(COutPoint(tx2a_txid, 0), nSequence=0xfffffffd)]
tx3a.vout = [CTxOut(int(0.9*COIN), CScript([b'c'])), CTxOut(int(0.9*COIN), CScript([b'd']))]
tx3a_hex = txToHex(tx3a)
tx3a_txid = self.nodes[0].sendrawtransaction(tx3a_hex, True)
# This transaction is shown as replaceable
assert_equal(self.nodes[0].getmempoolentry(tx3a_txid)['bip125-replaceable'], True)
tx3b = CTransaction()
tx3b.vin = [CTxIn(COutPoint(tx1a_txid, 0), nSequence=0)]
tx3b.vout = [CTxOut(int(0.5 * COIN), CScript([b'e' * 35]))]
tx3b_hex = txToHex(tx3b)
tx3c = CTransaction()
tx3c.vin = [CTxIn(COutPoint(tx2a_txid, 0), nSequence=0)]
tx3c.vout = [CTxOut(int(0.5 * COIN), CScript([b'f' * 35]))]
tx3c_hex = txToHex(tx3c)
self.nodes[0].sendrawtransaction(tx3b_hex, True)
# If tx3b was accepted, tx3c won't look like a replacement,
# but make sure it is accepted anyway
self.nodes[0].sendrawtransaction(tx3c_hex, True)
def mine_block(self,
node,
vtx=None,
mn_payee=None,
mn_amount=None,
use_mnmerkleroot_from_tip=False,
expected_error=None):
if vtx is None:
vtx = []
bt = node.getblocktemplate({'rules': ['segwit']})
height = bt['height']
tip_hash = bt['previousblockhash']
tip_block = node.getblock(tip_hash, 2)["tx"][0]
coinbasevalue = 50 * COIN
halvings = int(height / 150) # regtest
coinbasevalue >>= halvings
miner_script = self.nodes[0].getaddressinfo(
self.nodes[0].getnewaddress())['scriptPubKey']
if mn_payee is None:
if isinstance(bt['masternode'], list):
mn_payee = bt['masternode'][0]['script']
else:
mn_payee = bt['masternode']['script']
# we can't take the masternode payee amount from the template here as we might have additional fees in vtx
new_fees = 0
for tx in vtx:
in_value = 0
out_value = 0
for txin in tx.vin:
txout = node.gettxout("%064x" % txin.prevout.hash,
txin.prevout.n, False)
in_value += int(txout['value'] * COIN)
for txout in tx.vout:
out_value += txout.nValue
new_fees += in_value - out_value
if mn_amount is None:
mn_amount = get_masternode_payment(
height, coinbasevalue,
bt['masternode_collateral_height']) + new_fees / 2
miner_amount = int(coinbasevalue * 0.25)
miner_amount += new_fees / 2
coinbase = CTransaction()
coinbase.vout.append(
CTxOut(int(miner_amount), hex_str_to_bytes(miner_script)))
coinbase.vout.append(CTxOut(int(mn_amount),
hex_str_to_bytes(mn_payee)))
coinbase.vin = create_coinbase(height).vin
# Recreate mn root as using one in BT would result in invalid merkle roots for masternode lists
coinbase.nVersion = bt['version_coinbase']
if len(bt['default_witness_commitment_extra']) != 0:
if use_mnmerkleroot_from_tip:
cbtx = FromHex(CCbTx(version=2),
bt['default_witness_commitment_extra'])
if 'cbTx' in tip_block:
cbtx.merkleRootMNList = int(
tip_block['cbTx']['merkleRootMNList'], 16)
else:
cbtx.merkleRootMNList = 0
coinbase.extraData = cbtx.serialize()
else:
coinbase.extraData = hex_str_to_bytes(
bt['default_witness_commitment_extra'])
coinbase.calc_sha256(with_witness=True)
block = create_block(int(tip_hash, 16), coinbase)
block.nVersion = 4
block.vtx += vtx
block.hashMerkleRoot = block.calc_merkle_root()
add_witness_commitment(block)
block.solve()
result = node.submitblock(ToHex(block))
if expected_error is not None and result != expected_error:
raise AssertionError(
'mining the block should have failed with error %s, but submitblock returned %s'
% (expected_error, result))
elif expected_error is None and result is not None:
raise AssertionError('submitblock returned %s' % (result))
def test_sequence_lock_unconfirmed_inputs(self):
# Store height so we can easily reset the chain at the end of the test
cur_height = self.nodes[0].getblockcount()
# Create a mempool tx.
txid = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(), 2)
tx1 = FromHex(CTransaction(), self.nodes[0].getrawtransaction(txid))
tx1.rehash()
# Anyone-can-spend mempool tx.
# Sequence lock of 0 should pass.
tx2 = CTransaction()
tx2.nVersion = 2
tx2.vin = [CTxIn(COutPoint(tx1.sha256, 0), nSequence=0)]
tx2.vout = [
CTxOut(int(tx1.vout[0].nValue - self.relayfee * COIN),
CScript([b'a']))
]
tx2_raw = self.nodes[0].signrawtransactionwithwallet(ToHex(tx2))["hex"]
tx2 = FromHex(tx2, tx2_raw)
tx2.rehash()
self.nodes[0].sendrawtransaction(tx2_raw)
# Create a spend of the 0th output of orig_tx with a sequence lock
# of 1, and test what happens when submitting.
# orig_tx.vout[0] must be an anyone-can-spend output
def test_nonzero_locks(orig_tx, node, relayfee, use_height_lock):
sequence_value = 1
if not use_height_lock:
sequence_value |= SEQUENCE_LOCKTIME_TYPE_FLAG
tx = CTransaction()
tx.nVersion = 2
tx.vin = [
CTxIn(COutPoint(orig_tx.sha256, 0), nSequence=sequence_value)
]
tx.vout = [
CTxOut(int(orig_tx.vout[0].nValue - relayfee * COIN),
CScript([b'a' * 35]))
]
tx.rehash()
if (orig_tx.hash in node.getrawmempool()):
# sendrawtransaction should fail if the tx is in the mempool
assert_raises_rpc_error(-26, NOT_FINAL_ERROR,
node.sendrawtransaction, ToHex(tx))
else:
# sendrawtransaction should succeed if the tx is not in the mempool
node.sendrawtransaction(ToHex(tx))
return tx
test_nonzero_locks(tx2,
self.nodes[0],
self.relayfee,
use_height_lock=True)
test_nonzero_locks(tx2,
self.nodes[0],
self.relayfee,
use_height_lock=False)
# Now mine some blocks, but make sure tx2 doesn't get mined.
# Use prioritisetransaction to lower the effective feerate to 0
self.nodes[0].prioritisetransaction(txid=tx2.hash,
fee_delta=int(-self.relayfee *
COIN))
cur_time = int(time.time())
for i in range(10):
self.nodes[0].setmocktime(cur_time + 600)
self.nodes[0].generate(1)
cur_time += 600
assert (tx2.hash in self.nodes[0].getrawmempool())
test_nonzero_locks(tx2,
self.nodes[0],
self.relayfee,
use_height_lock=True)
test_nonzero_locks(tx2,
self.nodes[0],
self.relayfee,
use_height_lock=False)
# Mine tx2, and then try again
self.nodes[0].prioritisetransaction(txid=tx2.hash,
fee_delta=int(self.relayfee *
COIN))
# Advance the time on the node so that we can test timelocks
self.nodes[0].setmocktime(cur_time + 600)
self.nodes[0].generate(1)
assert (tx2.hash not in self.nodes[0].getrawmempool())
# Now that tx2 is not in the mempool, a sequence locked spend should
# succeed
tx3 = test_nonzero_locks(tx2,
self.nodes[0],
self.relayfee,
use_height_lock=False)
assert (tx3.hash in self.nodes[0].getrawmempool())
self.nodes[0].generate(1)
assert (tx3.hash not in self.nodes[0].getrawmempool())
# One more test, this time using height locks
tx4 = test_nonzero_locks(tx3,
self.nodes[0],
self.relayfee,
use_height_lock=True)
assert (tx4.hash in self.nodes[0].getrawmempool())
# Now try combining confirmed and unconfirmed inputs
tx5 = test_nonzero_locks(tx4,
self.nodes[0],
self.relayfee,
use_height_lock=True)
assert (tx5.hash not in self.nodes[0].getrawmempool())
utxos = self.nodes[0].listunspent()
tx5.vin.append(
CTxIn(COutPoint(int(utxos[0]["txid"], 16), utxos[0]["vout"]),
nSequence=1))
tx5.vout[0].nValue += int(utxos[0]["amount"] * COIN)
raw_tx5 = self.nodes[0].signrawtransactionwithwallet(ToHex(tx5))["hex"]
assert_raises_rpc_error(-26, NOT_FINAL_ERROR,
self.nodes[0].sendrawtransaction, raw_tx5)
# Test mempool-BIP68 consistency after reorg
#
# State of the transactions in the last blocks:
# ... -> [ tx2 ] -> [ tx3 ]
# tip-1 tip
# And currently tx4 is in the mempool.
#
# If we invalidate the tip, tx3 should get added to the mempool, causing
# tx4 to be removed (fails sequence-lock).
self.nodes[0].invalidateblock(self.nodes[0].getbestblockhash())
assert (tx4.hash not in self.nodes[0].getrawmempool())
assert (tx3.hash in self.nodes[0].getrawmempool())
# Now mine 2 empty blocks to reorg out the current tip (labeled tip-1 in
# diagram above).
# This would cause tx2 to be added back to the mempool, which in turn causes
# tx3 to be removed.
tip = int(
self.nodes[0].getblockhash(self.nodes[0].getblockcount() - 1), 16)
height = self.nodes[0].getblockcount()
for i in range(2):
block = create_block(tip, create_coinbase(height), cur_time)
block.nVersion = 3
block.rehash()
block.solve()
tip = block.sha256
height += 1
self.nodes[0].submitblock(ToHex(block))
cur_time += 1
mempool = self.nodes[0].getrawmempool()
assert (tx3.hash not in mempool)
assert (tx2.hash in mempool)
# Reset the chain and get rid of the mocktimed-blocks
self.nodes[0].setmocktime(0)
self.nodes[0].invalidateblock(self.nodes[0].getblockhash(cur_height +
1))
self.nodes[0].generate(10)
def test_sequence_lock_unconfirmed_inputs(self):
# Store height so we can easily reset the chain at the end of the test
cur_height = self.nodes[0].getblockcount()
# Create a mempool tx.
txid = self.nodes[0].sendtoaddress(self.nodes[0].getnewaddress(),
2000000)
tx1 = FromHex(CTransaction(), self.nodes[0].getrawtransaction(txid))
tx1.rehash()
# As the fees are calculated prior to the transaction being signed,
# there is some uncertainty that calculate fee provides the correct
# minimal fee. Since regtest coins are free, let's go ahead and
# increase the fee by an order of magnitude to ensure this test
# passes.
fee_multiplier = 10
# Anyone-can-spend mempool tx.
# Sequence lock of 0 should pass.
tx2 = CTransaction()
tx2.nVersion = 2
tx2.vin = [CTxIn(COutPoint(tx1.sha256, 0), nSequence=0)]
tx2.vout = [CTxOut(int(0), CScript([b'a']))]
tx2.vout[0].nValue = tx1.vout[0].nValue - \
fee_multiplier * self.nodes[0].calculate_fee(tx2)
tx2_raw = self.nodes[0].signrawtransactionwithwallet(ToHex(tx2))["hex"]
tx2 = FromHex(tx2, tx2_raw)
tx2.rehash()
self.nodes[0].sendrawtransaction(tx2_raw)
# Create a spend of the 0th output of orig_tx with a sequence lock
# of 1, and test what happens when submitting.
# orig_tx.vout[0] must be an anyone-can-spend output
def test_nonzero_locks(orig_tx, node, use_height_lock):
sequence_value = 1
if not use_height_lock:
sequence_value |= SEQUENCE_LOCKTIME_TYPE_FLAG
tx = CTransaction()
tx.nVersion = 2
tx.vin = [
CTxIn(COutPoint(orig_tx.sha256, 0), nSequence=sequence_value)
]
tx.vout = [
CTxOut(
int(orig_tx.vout[0].nValue -
fee_multiplier * node.calculate_fee(tx)),
CScript([b'a']))
]
pad_tx(tx)
tx.rehash()
if (orig_tx.hash in node.getrawmempool()):
# sendrawtransaction should fail if the tx is in the mempool
assert_raises_rpc_error(-26, NOT_FINAL_ERROR,
node.sendrawtransaction, ToHex(tx))
else:
# sendrawtransaction should succeed if the tx is not in the
# mempool
node.sendrawtransaction(ToHex(tx))
return tx
test_nonzero_locks(tx2, self.nodes[0], use_height_lock=True)
test_nonzero_locks(tx2, self.nodes[0], use_height_lock=False)
# Now mine some blocks, but make sure tx2 doesn't get mined.
# Use prioritisetransaction to lower the effective feerate to 0
self.nodes[0].prioritisetransaction(txid=tx2.hash,
fee_delta=-fee_multiplier *
self.nodes[0].calculate_fee(tx2))
cur_time = int(time.time())
for _ in range(10):
self.nodes[0].setmocktime(cur_time + 600)
self.nodes[0].generate(1)
cur_time += 600
assert tx2.hash in self.nodes[0].getrawmempool()
test_nonzero_locks(tx2, self.nodes[0], use_height_lock=True)
test_nonzero_locks(tx2, self.nodes[0], use_height_lock=False)
# Mine tx2, and then try again
self.nodes[0].prioritisetransaction(txid=tx2.hash,
fee_delta=fee_multiplier *
self.nodes[0].calculate_fee(tx2))
# Advance the time on the node so that we can test timelocks
self.nodes[0].setmocktime(cur_time + 600)
# Save block template now to use for the reorg later
tmpl = self.nodes[0].getblocktemplate()
self.nodes[0].generate(1)
assert tx2.hash not in self.nodes[0].getrawmempool()
# Now that tx2 is not in the mempool, a sequence locked spend should
# succeed
tx3 = test_nonzero_locks(tx2, self.nodes[0], use_height_lock=False)
assert tx3.hash in self.nodes[0].getrawmempool()
self.nodes[0].generate(1)
assert tx3.hash not in self.nodes[0].getrawmempool()
# One more test, this time using height locks
tx4 = test_nonzero_locks(tx3, self.nodes[0], use_height_lock=True)
assert tx4.hash in self.nodes[0].getrawmempool()
# Now try combining confirmed and unconfirmed inputs
tx5 = test_nonzero_locks(tx4, self.nodes[0], use_height_lock=True)
assert tx5.hash not in self.nodes[0].getrawmempool()
utxos = self.nodes[0].listunspent()
tx5.vin.append(
CTxIn(COutPoint(int(utxos[0]["txid"], 16), utxos[0]["vout"]),
nSequence=1))
tx5.vout[0].nValue += int(utxos[0]["amount"] * XEC)
raw_tx5 = self.nodes[0].signrawtransactionwithwallet(ToHex(tx5))["hex"]
assert_raises_rpc_error(-26, NOT_FINAL_ERROR,
self.nodes[0].sendrawtransaction, raw_tx5)
# Test mempool-BIP68 consistency after reorg
#
# State of the transactions in the last blocks:
# ... -> [ tx2 ] -> [ tx3 ]
# tip-1 tip
# And currently tx4 is in the mempool.
#
# If we invalidate the tip, tx3 should get added to the mempool, causing
# tx4 to be removed (fails sequence-lock).
self.nodes[0].invalidateblock(self.nodes[0].getbestblockhash())
assert tx4.hash not in self.nodes[0].getrawmempool()
assert tx3.hash in self.nodes[0].getrawmempool()
# Now mine 2 empty blocks to reorg out the current tip (labeled tip-1 in
# diagram above).
# This would cause tx2 to be added back to the mempool, which in turn causes
# tx3 to be removed.
for i in range(2):
block = create_block(tmpl=tmpl, ntime=cur_time)
block.rehash()
block.solve()
tip = block.sha256
assert_equal(None if i == 1 else 'inconclusive',
self.nodes[0].submitblock(ToHex(block)))
tmpl = self.nodes[0].getblocktemplate()
tmpl['previousblockhash'] = f"{tip:x}"
tmpl['transactions'] = []
cur_time += 1
mempool = self.nodes[0].getrawmempool()
assert tx3.hash not in mempool
assert tx2.hash in mempool
# Reset the chain and get rid of the mocktimed-blocks
self.nodes[0].setmocktime(0)
self.nodes[0].invalidateblock(self.nodes[0].getblockhash(cur_height +
1))
self.nodes[0].generate(10)
def create_self_transfer(self,
*,
fee_rate=Decimal("0.003"),
fee=Decimal("0"),
utxo_to_spend=None,
locktime=0,
sequence=0):
"""Create and return a tx with the specified fee. If fee is 0, use fee_rate, where the resulting fee may be exact or at most one satoshi higher than needed."""
utxo_to_spend = utxo_to_spend or self.get_utxo()
assert fee_rate >= 0
assert fee >= 0
if self._mode in (MiniWalletMode.RAW_OP_TRUE,
MiniWalletMode.ADDRESS_OP_TRUE):
vsize = Decimal(104) # anyone-can-spend
elif self._mode == MiniWalletMode.RAW_P2PK:
vsize = Decimal(
168
) # P2PK (73 bytes scriptSig + 35 bytes scriptPubKey + 60 bytes other)
else:
assert False
send_value = utxo_to_spend["value"] - (fee or
(fee_rate * vsize / 1000))
assert send_value > 0
tx = CTransaction()
tx.vin = [
CTxIn(COutPoint(int(utxo_to_spend['txid'], 16),
utxo_to_spend['vout']),
nSequence=sequence)
]
tx.vout = [
CTxOut(int(COIN * send_value), bytearray(self._scriptPubKey))
]
tx.nLockTime = locktime
if self._mode == MiniWalletMode.RAW_P2PK:
self.sign_tx(tx)
elif self._mode == MiniWalletMode.RAW_OP_TRUE:
tx.vin[0].scriptSig = CScript([OP_NOP] *
43) # pad to identical size
elif self._mode == MiniWalletMode.ADDRESS_OP_TRUE:
tx.wit.vtxinwit = [CTxInWitness()]
tx.wit.vtxinwit[0].scriptWitness.stack = [
CScript([OP_TRUE]),
bytes([LEAF_VERSION_TAPSCRIPT]) + self._internal_key
]
else:
assert False
tx_hex = tx.serialize().hex()
assert_equal(tx.get_vsize(), vsize)
new_utxo = self._create_utxo(txid=tx.rehash(),
vout=0,
value=send_value,
height=0)
return {
"txid": new_utxo["txid"],
"wtxid": tx.getwtxid(),
"hex": tx_hex,
"tx": tx,
"new_utxo": new_utxo
}
def SignatureHash_legacy(script, txTo, inIdx, hashtype):
"""
This method is identical to the regular `SignatureHash` method,
but without support for SIGHASH_RANGEPROOF.
So basically it's the old version of the method from before the
new sighash flag was added.
"""
HASH_ONE = b'\x01\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
if inIdx >= len(txTo.vin):
return (HASH_ONE,
"inIdx %d out of range (%d)" % (inIdx, len(txTo.vin)))
txtmp = CTransaction(txTo)
for txin in txtmp.vin:
txin.scriptSig = b''
txtmp.vin[inIdx].scriptSig = FindAndDelete(script,
CScript([OP_CODESEPARATOR]))
if (hashtype & 0x1f) == SIGHASH_NONE:
txtmp.vout = []
for i in range(len(txtmp.vin)):
if i != inIdx:
txtmp.vin[i].nSequence = 0
elif (hashtype & 0x1f) == SIGHASH_SINGLE:
outIdx = inIdx
if outIdx >= len(txtmp.vout):
return (HASH_ONE,
"outIdx %d out of range (%d)" % (outIdx, len(txtmp.vout)))
tmp = txtmp.vout[outIdx]
txtmp.vout = []
for i in range(outIdx):
txtmp.vout.append(CTxOut(-1))
txtmp.vout.append(tmp)
for i in range(len(txtmp.vin)):
if i != inIdx:
txtmp.vin[i].nSequence = 0
if hashtype & SIGHASH_ANYONECANPAY:
tmp = txtmp.vin[inIdx]
txtmp.vin = []
txtmp.vin.append(tmp)
# sighash serialization is different from non-witness serialization
# do manual sighash serialization:
s = b""
s += struct.pack("<i", txtmp.nVersion)
s += ser_vector(txtmp.vin)
s += ser_vector(txtmp.vout)
s += struct.pack("<I", txtmp.nLockTime)
# add sighash type
s += struct.pack(b"<I", hashtype)
hash = hash256(s)
return (hash, None)
