def test_extreme_compute_jumps_convergence(self):
"""Test of compute_jumps() requiring absurd number of passes.
NOTE: This test also serves to demonstrate that there is no worst
case: the number of passes can be unlimited (or, actually, limited by
the size of the provided code).
This is an extension of test_compute_jumps_convergence. Instead of
two jumps, where the earlier gets extended after the latter, we
instead generate a series of many jumps. Each pass of compute_jumps()
extends one more instruction, which in turn causes the one behind it
to be extended on the next pass.
"""
if not WORDCODE:
return
# N: the number of unextended instructions that can be squeezed into a
# set of bytes adressable by the arg of an unextended instruction.
# The answer is "128", but here's how we arrive at it (and it also
# hints at how to make this work for pre-WORDCODE).
max_unextended_offset = 1 << 8
unextended_branch_instr_size = 2
N = max_unextended_offset // unextended_branch_instr_size
nop = "UNARY_POSITIVE" # don't use NOP, dis.stack_effect will raise
# The number of jumps will be equal to the number of labels. The
# number of passes of compute_jumps() required will be one greater
# than this.
labels = [Label() for x in range(0, 3 * N)]
code = Bytecode()
code.extend(
Instr("JUMP_FORWARD", labels[len(labels) - x - 1])
for x in range(0, len(labels))
)
end_of_jumps = len(code)
code.extend(Instr(nop) for x in range(0, N))
# Now insert the labels. The first is N instructions (i.e. 256
# bytes) after the last jump. Then they proceed to earlier positions
# 4 bytes at a time. While the targets are in the range of the nop
# instructions, 4 bytes is two instructions. When the targets are in
# the range of JUMP_FORWARD instructions we have to allow for the fact
# that the instructions will have been extended to four bytes each, so
# working backwards 4 bytes per label means just one instruction per
# label.
offset = end_of_jumps + N
for index in range(0, len(labels)):
code.insert(offset, labels[index])
if offset <= end_of_jumps:
offset -= 1
else:
offset -= 2
code.insert(0, Instr("LOAD_CONST", 0))
del end_of_jumps
code.append(Instr("RETURN_VALUE"))
code.to_code(compute_jumps_passes=(len(labels) + 1))
def test_invalid_types(self):
code = Bytecode()
code.append(123)
with self.assertRaises(ValueError):
list(code)
with self.assertRaises(ValueError):
code.legalize()
with self.assertRaises(ValueError):
Bytecode([123])
def test_compute_jumps_convergence(self):
# Consider the following sequence of instructions:
#
# JUMP_ABSOLUTE Label1
# JUMP_ABSOLUTE Label2
# ...126 instructions...
# Label1: Offset 254 on first pass, 256 second pass
# NOP
# ... many more instructions ...
# Label2: Offset > 256 on first pass
#
# On first pass of compute_jumps(), Label2 will be at address 254, so
# that value encodes into the single byte arg of JUMP_ABSOLUTE.
#
# On second pass compute_jumps() the instr at Label1 will have offset
# of 256 so will also be given an EXTENDED_ARG.
#
# Thus we need to make an additional pass. This test only verifies
# case where 2 passes is insufficient but three is enough.
#
# On Python > 3.10 we need to double the number since the offset is now
# in term of instructions and not bytes.
# Create code from comment above.
code = Bytecode()
label1 = Label()
label2 = Label()
nop = "NOP"
code.append(Instr("JUMP_ABSOLUTE", label1))
code.append(Instr("JUMP_ABSOLUTE", label2))
# Need 254 * 2 + 2 since the arg will change by 1 instruction rather than 2
# bytes.
for x in range(4, 510 if OFFSET_AS_INSTRUCTION else 254, 2):
code.append(Instr(nop))
code.append(label1)
code.append(Instr(nop))
for x in range(
514 if OFFSET_AS_INSTRUCTION else 256,
600 if OFFSET_AS_INSTRUCTION else 300,
2,
):
code.append(Instr(nop))
code.append(label2)
code.append(Instr(nop))
# This should pass by default.
code.to_code()
# Try with max of two passes: it should raise
with self.assertRaises(RuntimeError):
code.to_code(compute_jumps_passes=2)
def test_compute_jumps_convergence(self):
# Consider the following sequence of instructions:
#
# JUMP_ABSOLUTE Label1
# JUMP_ABSOLUTE Label2
# ...126 instructions...
# Label1: Offset 254 on first pass, 256 second pass
# NOP
# ... many more instructions ...
# Label2: Offset > 256 on first pass
#
# On first pass of compute_jumps(), Label2 will be at address 254, so
# that value encodes into the single byte arg of JUMP_ABSOLUTE.
#
# On second pass compute_jumps() the instr at Label1 will have offset
# of 256 so will also be given an EXTENDED_ARG.
#
# Thus we need to make an additional pass. This test only verifies
# case where 2 passes is insufficient but three is enough.
if not WORDCODE:
# Could be done pre-WORDCODE, but that requires 2**16 bytes of
# code.
return
# Create code from comment above.
code = Bytecode()
label1 = Label()
label2 = Label()
nop = "NOP"
code.append(Instr("JUMP_ABSOLUTE", label1))
code.append(Instr("JUMP_ABSOLUTE", label2))
for x in range(4, 254, 2):
code.append(Instr(nop))
code.append(label1)
code.append(Instr(nop))
for x in range(256, 300, 2):
code.append(Instr(nop))
code.append(label2)
code.append(Instr(nop))
# This should pass by default.
code.to_code()
# Try with max of two passes: it should raise
with self.assertRaises(RuntimeError):
code.to_code(compute_jumps_passes=2)