如何检查用exec创建的函数代码

Question

I have a function that was created with exec() inside another function and passed as an argument to the main program. How can I get the code of said function? I tried inspec.getsourcelines() and inspec.getsource() but I get the following error

IOError: could not get source code

Is there a way around this?

MWE main file:

#!/usr/bin/python
from ext import makeF
import inspect

f=makeF()
f()
inspect.getsource(f)

then external file:

def makeF():
    script="def sayA():\n\tprint('Aah')"
    exec(script)
    return sayA

Answer 1

This is not possible. I've been digging around and came to the same conclusion as outlined in this answer .

I don't know your code, but I think in your specific case you could return a custom object containing the source code of the function (which you seem to have - you're passing it to exec) together with the actual compiled function. You could even leverage Python's __call__ magic method to better emulate the original behaviour. Here's a sample:

class FunctionWithCode:
    def __init__(self, source, func_name):
        exec(source)
        self.source = source
        self.func = locals()[func_name]

    def __call__(self, *args, **kwargs):
        self.func(*args, **kwargs)

f = FunctionWithCode("def sayA():\n\tprint('Aah')", 'sayA')

f()

which prints Aah as expected. You would need to know the function name when creating the object but this is the same for your sample code.

Answer 2

In python 3 the solution is the __globals__ attribute of the functions.

With your example :

>>> f.__globals__['txt']
"def sayA():\n\tprint('Aah')"

Unfortunately I could not find anything like for Python 2.

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The reason why the other methods cannot work, is because they use the filename of the module and here you do not have any file.