New submission from Peter Ebden <peter.eb...@gmail.com>:
We've found that the following code produces non-deterministic bytecode,
even post PEP-552:
def test(x):
if x in {'ONE', 'TWO', 'THREE'}:
pass
It's not too hard to test it:
$ python3.7 -m compileall --invalidation-mode=unchecked-hash test.py
Compiling 'test.py'...
$ sha1sum __pycache__/test.cpython-37.pyc
61e5682ca95e8707b4ef2a79f64566664dafd800 __pycache__/test.cpython-37.pyc
$ rm __pycache__/test.cpython-37.pyc
$ python3.7 -m compileall --invalidation-mode=unchecked-hash test.py
Compiling 'test.py'...
$ sha1sum __pycache__/test.cpython-37.pyc
222a06621b491879e5317b34e9dd715bacd89b7d __pycache__/test.cpython-37.pyc
It looks like the peephole optimiser is converting the LOAD_CONST instructions
for the set into a single LOAD_CONST for a frozenset which then serialises in
nondeterministic order. One can hence work around it by setting PYTHONHASHSEED
to a known value.
I'm happy to help out with this if needed, although I don't have a lot of
familiarity with the relevant code.
----------
components: Interpreter Core
messages: 325644
nosy: Peter Ebden
priority: normal
severity: normal
status: open
title: Non-deterministic bytecode generation
versions: Python 3.7
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue34722>
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