If I have a reference to a function I can check it\'s code object f.__code__
, get a signature, then perform later checks against this signature to see if the code c
You may to compare bytecode attributes on code object using method.__code__.co_code
. For example lets define two classes:
>>> class A:
... a = 1
... def b(self, b):
... print(self.a + b)
...
>>> class B:
... a = 1
... def b(self, b):
... print(self.a + b)
...
>>> A().b.__code__.co_code
'|\x00\x00j\x00\x00|\x01\x00\x17GHd\x00\x00S'
>>> B().b.__code__.co_code
'|\x00\x00j\x00\x00|\x01\x00\x17GHd\x00\x00S'
>>> A().b.__code__.co_code == B().b.__code__.co_code
True
and if method b
in class A
is changed:
>>> class A:
... a = 1
... def b(self, b):
... print(b + self.a)
...
>>> A().b.__code__.co_code
'|\x01\x00|\x00\x00j\x00\x00\x17GHd\x00\x00S'
>>> A().b.__code__.co_code == B().b.__code__.co_code
False
or use inspect method inspect.getsource(object) that:
Return the text of the source code for an object. The argument may be a module, class, method, function, traceback, frame, or code object. The source code is returned as a single string.
And if you want to know whether the code has changed in dynamic you may need to reload your class with importlib and compare bytecode.
I'm just poking around at the __code__
object here, so I can't say that this would work for sure, but it looks to me like you could use the co_names
tuple to (recursively) traverse the call graph rooted at a particular function, in order to build up some sort of hash of the transitive closure of the functions that could possible be called. (I don't think it'd be possible to include only the functions that will be called for a particular input, but you could be conservative and include every possible call.)
To do this you'd need to maintain some sort of symbol table to be able to look up the names of the functions that get called. But once you start going down this path, it seems like you're basically going to build up your own equivalent of the AST. So, why not just use the AST to start with ?