def run_test (self):
# Mine some coins
self.nodes[0].generate(110)
# Get some addresses from the two nodes
addr1 = [self.nodes[1].getnewaddress() for i in range(3)]
addr2 = [self.nodes[2].getnewaddress() for i in range(3)]
addrs = addr1 + addr2
# Send 1 + 0.5 coin to each address
[self.nodes[0].sendtoaddress(addr, 1.0) for addr in addrs]
[self.nodes[0].sendtoaddress(addr, 0.5) for addr in addrs]
self.nodes[0].generate(1)
self.sync_all()
# For each node, send 0.2 coins back to 0;
# - node[1] should pick one 0.5 UTXO and leave the rest
# - node[2] should pick one (1.0 + 0.5) UTXO group corresponding to a
# given address, and leave the rest
txid1 = self.nodes[1].sendtoaddress(self.nodes[0].getnewaddress(), 0.2)
tx1 = self.nodes[1].getrawtransaction(txid1, True)
# txid1 should have 1 input and 2 outputs
assert_equal(1, len(tx1["vin"]))
assert_equal(2, len(tx1["vout"]))
# one output should be 0.2, the other should be ~0.3
v = [vout["value"] for vout in tx1["vout"]]
v.sort()
assert_approx(v[0], 0.2)
assert_approx(v[1], 0.3, 0.0001)
txid2 = self.nodes[2].sendtoaddress(self.nodes[0].getnewaddress(), 0.2)
tx2 = self.nodes[2].getrawtransaction(txid2, True)
# txid2 should have 2 inputs and 2 outputs
assert_equal(2, len(tx2["vin"]))
assert_equal(2, len(tx2["vout"]))
# one output should be 0.2, the other should be ~1.3
v = [vout["value"] for vout in tx2["vout"]]
v.sort()
assert_approx(v[0], 0.2)
assert_approx(v[1], 1.3, 0.0001)
# Empty out node2's wallet
self.nodes[2].sendtoaddress(address=self.nodes[0].getnewaddress(), amount=self.nodes[2].getbalance(), subtractfeefromamount=True)
self.sync_all()
self.nodes[0].generate(1)
# Fill node2's wallet with 10000 outputs corresponding to the same
# scriptPubKey
for i in range(5):
raw_tx = self.nodes[0].createrawtransaction([{"txid":"0"*64, "vout":0}], [{addr2[0]: 0.05}])
tx = FromHex(CTransaction(), raw_tx)
tx.vin = []
tx.vout = [tx.vout[0]] * 2000
funded_tx = self.nodes[0].fundrawtransaction(ToHex(tx))
signed_tx = self.nodes[0].signrawtransactionwithwallet(funded_tx['hex'])
self.nodes[0].sendrawtransaction(signed_tx['hex'])
self.nodes[0].generate(1)
self.sync_all()
# Check that we can create a transaction that only requires ~100 of our
# utxos, without pulling in all outputs and creating a transaction that
# is way too big.
assert self.nodes[2].sendtoaddress(address=addr2[0], amount=5)
def run_test(self):
# Mine some coins
self.nodes[0].generate(110)
# Get some addresses from the two nodes
addr1 = [self.nodes[1].getnewaddress() for _ in range(3)]
addr2 = [self.nodes[2].getnewaddress() for _ in range(3)]
addrs = addr1 + addr2
# Send 1 + 0.5 coin to each address
[self.nodes[0].sendtoaddress(addr, 1.0) for addr in addrs]
[self.nodes[0].sendtoaddress(addr, 0.5) for addr in addrs]
self.nodes[0].generate(1)
self.sync_all()
# For each node, send 0.2 coins back to 0;
# - node[1] should pick one 0.5 UTXO and leave the rest
# - node[2] should pick one (1.0 + 0.5) UTXO group corresponding to a
# given address, and leave the rest
txid1 = self.nodes[1].sendtoaddress(self.nodes[0].getnewaddress(), 0.2)
tx1 = self.nodes[1].getrawtransaction(txid1, True)
# txid1 should have 1 input and 2 outputs
assert_equal(1, len(tx1["vin"]))
assert_equal(2, len(tx1["vout"]))
# one output should be 0.2, the other should be ~0.3
v = [vout["value"] for vout in tx1["vout"]]
v.sort()
assert_approx(v[0], vexp=0.2, vspan=0.0001)
assert_approx(v[1], vexp=0.3, vspan=0.0001)
txid2 = self.nodes[2].sendtoaddress(self.nodes[0].getnewaddress(), 0.2)
tx2 = self.nodes[2].getrawtransaction(txid2, True)
# txid2 should have 2 inputs and 2 outputs
assert_equal(2, len(tx2["vin"]))
assert_equal(2, len(tx2["vout"]))
# one output should be 0.2, the other should be ~1.3
v = [vout["value"] for vout in tx2["vout"]]
v.sort()
assert_approx(v[0], vexp=0.2, vspan=0.0001)
assert_approx(v[1], vexp=1.3, vspan=0.0001)
# Test 'avoid partial if warranted, even if disabled'
self.sync_all()
self.nodes[0].generate(1)
# Nodes 1-2 now have confirmed UTXOs (letters denote destinations):
# Node #1: Node #2:
# - A 1.0 - D0 1.0
# - B0 1.0 - D1 0.5
# - B1 0.5 - E0 1.0
# - C0 1.0 - E1 0.5
# - C1 0.5 - F ~1.3
# - D ~0.3
assert_approx(self.nodes[1].getbalance(), vexp=4.3, vspan=0.0001)
assert_approx(self.nodes[2].getbalance(), vexp=4.3, vspan=0.0001)
# Sending 1.4 afco should pick one 1.0 + one more. For node #1,
# this could be (A / B0 / C0) + (B1 / C1 / D). We ensure that it is
# B0 + B1 or C0 + C1, because this avoids partial spends while not being
# detrimental to transaction cost
txid3 = self.nodes[1].sendtoaddress(self.nodes[0].getnewaddress(), 1.4)
tx3 = self.nodes[1].getrawtransaction(txid3, True)
# tx3 should have 2 inputs and 2 outputs
assert_equal(2, len(tx3["vin"]))
assert_equal(2, len(tx3["vout"]))
# the accumulated value should be 1.5, so the outputs should be
# ~0.1 and 1.4 and should come from the same destination
values = [vout["value"] for vout in tx3["vout"]]
values.sort()
assert_approx(values[0], vexp=0.1, vspan=0.0001)
assert_approx(values[1], vexp=1.4, vspan=0.0001)
input_txids = [vin["txid"] for vin in tx3["vin"]]
input_addrs = [
self.nodes[1].gettransaction(txid)['details'][0]['address']
for txid in input_txids
]
assert_equal(input_addrs[0], input_addrs[1])
# Node 2 enforces avoidpartialspends so needs no checking here
# Test wallet option maxapsfee with Node 3
addr_aps = self.nodes[3].getnewaddress()
self.nodes[0].sendtoaddress(addr_aps, 1.0)
self.nodes[0].sendtoaddress(addr_aps, 1.0)
self.nodes[0].generate(1)
self.sync_all()
with self.nodes[3].assert_debug_log(
['Fee non-grouped = 2820, grouped = 4160, using grouped']):
txid4 = self.nodes[3].sendtoaddress(self.nodes[0].getnewaddress(),
0.1)
tx4 = self.nodes[3].getrawtransaction(txid4, True)
# tx4 should have 2 inputs and 2 outputs although one output would
# have been enough and the transaction caused higher fees
assert_equal(2, len(tx4["vin"]))
assert_equal(2, len(tx4["vout"]))
addr_aps2 = self.nodes[3].getnewaddress()
[self.nodes[0].sendtoaddress(addr_aps2, 1.0) for _ in range(5)]
self.nodes[0].generate(1)
self.sync_all()
with self.nodes[3].assert_debug_log(
['Fee non-grouped = 5520, grouped = 8240, using non-grouped']):
txid5 = self.nodes[3].sendtoaddress(self.nodes[0].getnewaddress(),
2.95)
tx5 = self.nodes[3].getrawtransaction(txid5, True)
# tx5 should have 3 inputs (1.0, 1.0, 1.0) and 2 outputs
assert_equal(3, len(tx5["vin"]))
assert_equal(2, len(tx5["vout"]))
# Test wallet option maxapsfee with node 4, which sets maxapsfee
# 1 sat higher, crossing the threshold from non-grouped to grouped.
addr_aps3 = self.nodes[4].getnewaddress()
[self.nodes[0].sendtoaddress(addr_aps3, 1.0) for _ in range(5)]
self.nodes[0].generate(1)
self.sync_all()
with self.nodes[4].assert_debug_log(
['Fee non-grouped = 5520, grouped = 8240, using grouped']):
txid6 = self.nodes[4].sendtoaddress(self.nodes[0].getnewaddress(),
2.95)
tx6 = self.nodes[4].getrawtransaction(txid6, True)
# tx6 should have 5 inputs and 2 outputs
assert_equal(5, len(tx6["vin"]))
assert_equal(2, len(tx6["vout"]))
# Empty out node2's wallet
self.nodes[2].sendtoaddress(address=self.nodes[0].getnewaddress(),
amount=self.nodes[2].getbalance(),
subtractfeefromamount=True)
self.sync_all()
self.nodes[0].generate(1)
# Fill node2's wallet with 10000 outputs corresponding to the same
# scriptPubKey
for _ in range(5):
raw_tx = self.nodes[0].createrawtransaction([{
"txid": "0" * 64,
"vout": 0
}], [{
addr2[0]: 0.05
}])
tx = FromHex(CTransaction(), raw_tx)
tx.vin = []
tx.vout = [tx.vout[0]] * 2000
funded_tx = self.nodes[0].fundrawtransaction(ToHex(tx))
signed_tx = self.nodes[0].signrawtransactionwithwallet(
funded_tx['hex'])
self.nodes[0].sendrawtransaction(signed_tx['hex'])
self.nodes[0].generate(1)
self.sync_all()
# Check that we can create a transaction that only requires ~100 of our
# utxos, without pulling in all outputs and creating a transaction that
# is way too big.
assert self.nodes[2].sendtoaddress(address=addr2[0], amount=5)
def run_test(self):
# Mine some coins
self.nodes[0].generate(110)
# Get some addresses from the two nodes
addr1 = [self.nodes[1].getnewaddress() for i in range(3)]
addr2 = [self.nodes[2].getnewaddress() for i in range(3)]
addrs = addr1 + addr2
# Send 1 + 0.5 coin to each address
[self.nodes[0].sendtoaddress(addr, 1.0) for addr in addrs]
[self.nodes[0].sendtoaddress(addr, 0.5) for addr in addrs]
self.nodes[0].generate(1)
self.sync_all()
# For each node, send 0.2 coins back to 0;
# - node[1] should pick one 0.5 UTXO and leave the rest
# - node[2] should pick one (1.0 + 0.5) UTXO group corresponding to a
# given address, and leave the rest
txid1 = self.nodes[1].sendtoaddress(self.nodes[0].getnewaddress(), 0.2)
tx1 = self.nodes[1].getrawtransaction(txid1, True)
# txid1 should have 1 input and 2 outputs
assert_equal(1, len(tx1["vin"]))
assert_equal(2, len(tx1["vout"]))
# one output should be 0.2, the other should be ~0.3
v = [vout["value"] for vout in tx1["vout"]]
v.sort()
assert_approx(v[0], 0.2)
assert_approx(v[1], 0.3, 0.0001)
txid2 = self.nodes[2].sendtoaddress(self.nodes[0].getnewaddress(), 0.2)
tx2 = self.nodes[2].getrawtransaction(txid2, True)
# txid2 should have 2 inputs and 2 outputs
assert_equal(2, len(tx2["vin"]))
assert_equal(2, len(tx2["vout"]))
# one output should be 0.2, the other should be ~1.3
v = [vout["value"] for vout in tx2["vout"]]
v.sort()
assert_approx(v[0], 0.2)
assert_approx(v[1], 1.3, 0.0001)
# Empty out node2's wallet
self.nodes[2].sendtoaddress(address=self.nodes[0].getnewaddress(), amount=self.nodes[2].getbalance(), subtractfeefromamount=True)
self.sync_all()
self.nodes[0].generate(1)
# Fill node2's wallet with 10000 outputs corresponding to the same
# scriptPubKey
for i in range(5):
raw_tx = self.nodes[0].createrawtransaction([{"txid":"0"*64, "vout":0}], [{addr2[0]: 0.05}])
tx = FromHex(CTransaction(), raw_tx)
tx.vin = []
tx.vout = [tx.vout[0]] * 2000
funded_tx = self.nodes[0].fundrawtransaction(ToHex(tx))
signed_tx = self.nodes[0].signrawtransactionwithwallet(funded_tx['hex'])
self.nodes[0].sendrawtransaction(signed_tx['hex'])
self.nodes[0].generate(1)
self.sync_all()
# Check that we can create a transaction that only requires ~100 of our
# utxos, without pulling in all outputs and creating a transaction that
# is way too big.
assert self.nodes[2].sendtoaddress(address=addr2[0], amount=5)
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')
# Pick a random coin(base) to spend
coin = coins.pop()
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 = 0.00000700
raw_tx_0 = node.signrawtransactionwithwallet(node.createrawtransaction(
inputs=[{"txid": txid_in_block, "vout": 0,
"sequence": 0xfffffffd}],
outputs=[{node.getnewaddress(): 0.3 - fee}],
))['hex']
tx = FromHex(CTransaction(), 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')
# Pick a random coin(base) to spend
coin = coins.pop()
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 = FromHex(CTransaction(), 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': 'txn-already-in-mempool'}],
rawtxs=[raw_tx_0],
)
# Removed RBF test
# self.log.info('A transaction that replaces a mempool transaction')
# ...
self.log.info('A transaction that conflicts with an unconfirmed tx')
# Send the transaction that conflicts with the mempool transaction
node.sendrawtransaction(hexstring=tx.serialize().hex(), maxfeerate=0)
# take original raw_tx_0
tx = FromHex(CTransaction(), 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 = FromHex(CTransaction(), 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=[ToHex(tx)],
)
self.log.info(
'A transaction with missing inputs, that existed once in the past')
tx = FromHex(CTransaction(), raw_tx_0)
# Set vout to 1, to spend the other outpoint (49 coins) of the
# in-chain-tx we want to double spend
tx.vin[0].prevout.n = 1
raw_tx_1 = node.signrawtransactionwithwallet(
ToHex(tx))['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 = FromHex(CTransaction(), raw_tx_reference)
# Reference tx should be valid on itself
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(), 'allowed': True}],
rawtxs=[ToHex(tx)],
maxfeerate=0,
)
self.log.info('A transaction with no outputs')
tx = FromHex(CTransaction(), raw_tx_reference)
tx.vout = []
# Skip re-signing the transaction for context independent checks from now on
# FromHex(tx, node.signrawtransactionwithwallet(ToHex(tx))['hex'])
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(
), 'allowed': False, 'reject-reason': 'bad-txns-vout-empty'}],
rawtxs=[ToHex(tx)],
)
self.log.info('A really large transaction')
tx = FromHex(CTransaction(), raw_tx_reference)
tx.vin = [tx.vin[0]] * (1 + 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=[ToHex(tx)],
)
self.log.info('A transaction with negative output value')
tx = FromHex(CTransaction(), 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=[ToHex(tx)],
)
# 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 = FromHex(CTransaction(), raw_tx_reference)
tx.vout[0].nValue = 21000000 * COIN + 1
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(
), 'allowed': False, 'reject-reason': 'bad-txns-vout-toolarge'}],
rawtxs=[ToHex(tx)],
)
self.log.info('A transaction with too large sum of output values')
tx = FromHex(CTransaction(), 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': 'bad-txns-txouttotal-toolarge'}],
rawtxs=[ToHex(tx)],
)
self.log.info('A transaction with duplicate inputs')
tx = FromHex(CTransaction(), 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=[ToHex(tx)],
)
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 = FromHex(CTransaction(), raw_tx_coinbase_spent)
self.check_mempool_result(
result_expected=[
{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-tx-coinbase'}],
rawtxs=[ToHex(tx)],
)
self.log.info('Some nonstandard transactions')
tx = FromHex(CTransaction(), 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=[ToHex(tx)],
)
tx = FromHex(CTransaction(), 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=[ToHex(tx)],
)
tx = FromHex(CTransaction(), raw_tx_reference)
# Some not-pushonly scriptSig
tx.vin[0].scriptSig = CScript([OP_HASH160])
self.check_mempool_result(
result_expected=[{'txid': tx.rehash(
), 'allowed': False, 'reject-reason': 'scriptsig-not-pushonly'}],
rawtxs=[ToHex(tx)],
)
tx = FromHex(CTransaction(), raw_tx_reference)
output_p2sh_burn = CTxOut(nValue=540, scriptPubKey=CScript(
[OP_HASH160, hash160(b'burn'), OP_EQUAL]))
# Use enough outputs to make the tx too large for our policy
num_scripts = 100000 // len(output_p2sh_burn.serialize())
tx.vout = [output_p2sh_burn] * num_scripts
self.check_mempool_result(
result_expected=[
{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'tx-size'}],
rawtxs=[ToHex(tx)],
)
tx = FromHex(CTransaction(), raw_tx_reference)
tx.vout[0] = output_p2sh_burn
# Make output smaller, such that it is dust for our policy
tx.vout[0].nValue -= 1
self.check_mempool_result(
result_expected=[
{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'dust'}],
rawtxs=[ToHex(tx)],
)
tx = FromHex(CTransaction(), 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': 'multi-op-return'}],
rawtxs=[ToHex(tx)],
)
self.log.info('A timelocked transaction')
tx = FromHex(CTransaction(), raw_tx_reference)
# Should be non-max, so locktime is not ignored
tx.vin[0].nSequence -= 1
tx.nLockTime = node.getblockcount() + 1
self.check_mempool_result(
result_expected=[
{'txid': tx.rehash(), 'allowed': False, 'reject-reason': 'bad-txns-nonfinal'}],
rawtxs=[ToHex(tx)],
)
self.log.info('A transaction that is locked by BIP68 sequence logic')
tx = FromHex(CTransaction(), raw_tx_reference)
# We could include it in the second block mined from now, but not the
# very next one
tx.vin[0].nSequence = 2
# 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,
)
assert_equal(2, len(tx6["vout"]))
# Empty out node2's wallet
self.nodes[2].sendtoaddress(address=self.nodes[0].getnewaddress(
), amount=self.nodes[2].getbalance(), subtractfeefromamount=True)
self.sync_all()
self.nodes[0].generate(1)
# Fill node2's wallet with 10000 outputs corresponding to the same
# scriptPubKey
for _ in range(5):
raw_tx = self.nodes[0].createrawtransaction(
[{"txid": "0" * 64, "vout": 0}], [{addr2[0]: 50_000}])
tx = FromHex(CTransaction(), raw_tx)
tx.vin = []
tx.vout = [tx.vout[0]] * 2000
funded_tx = self.nodes[0].fundrawtransaction(ToHex(tx))
signed_tx = self.nodes[0].signrawtransactionwithwallet(
funded_tx['hex'])
self.nodes[0].sendrawtransaction(signed_tx['hex'])
self.nodes[0].generate(1)
self.sync_all()
# Check that we can create a transaction that only requires ~100 of our
# utxos, without pulling in all outputs and creating a transaction that
# is way too big.
assert self.nodes[2].sendtoaddress(address=addr2[0], amount=5000000)
if __name__ == '__main__':
