Python 3 bytecode format

Question

I want to read a .pyc file. However, I cannot find any documentation on the format.

The only one I found does not work for Python 3 (although it does for Python 2):

>>> f = open('__pycache__/foo.cpython-34.pyc', 'rb') >>> f.read(4) b'\xee\x0c\r\n' >>> f.read(4) b'\xf8\x17\x08W' >>> marshal.load(f) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: bad marshal data (unknown type code)

marshal only consumes one byte: \x00, which indeed is not a valid first character for marshall (as a comparison, the first byte of Python 2 bytecode for the same empty module is c)

So, how can I decode what comes after the header?

From version Pytho3.3+ you need read not 8 first bytes, but 12. Read my answer here stackoverflow.com/a/42720524/6003870 — ВелоКастръ
– ВелоКастръ, Commented Mar 10, 2017 at 14:18

aghast · Accepted Answer · 2016-04-08 21:19:00Z

Try this. It worked a while back. They added another int32 in v3.

def load_file(self, source): if isinstance(source, str): import os.path if not os.path.exists(source): raise IOError("Cannot load_file('" + source + "'): does not exist") with open(source, "rb") as fh: header_bytes = fh.read(12) # ignore header self.code = marshal.load(fh) return self.code

This seems as if it should be a normal function (without self).

Robert Jacobs · Accepted Answer · 2016-04-08 21:14:45Z

0

Have you looked at the dissembler? https://docs.python.org/3/library/dis.html

answered Apr 8, 2016 at 21:14

Robert Jacobs

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1 Comment

Valentin Lorentz Over a year ago

The disassembler takes a code object, which must be decoded (the resource I linked uses marshal for that)

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