How do I get the return value when using Python exec on the code object of a function?

Question

For testing purposes I want to directly execute a function defined inside of another function.

I can get to the code object of the child function, through the code (func_code) of the parent function, but when I exec it, i get no return value.

Is there a way to get the return value from the exec'ed code?

Solution 1

Yes, you need to have the assignment within the exec statement:

>>> def foo():
...     return 5
...
>>> exec("a = foo()")
>>> a
5

This probably isn't relevant for your case since its being used in controlled testing, but be careful with using exec with user defined input.

Solution 2

A few years later, but the following snippet helped me:

the_code = '''
a = 1
b = 2
return_me = a + b
'''

loc = {}
exec(the_code, globals(), loc)
return_workaround = loc['return_me']
print(return_workaround)  # 3

exec() doesn't return anything itself, but you can pass a dict which has all the local variables stored in it after execution. By accessing it you have a something like a return.

I hope it helps someone.

Solution 3

While this is the ugliest beast ever seen by mankind, this is how you can do it by using a global variable inside your exec call:

def my_exec(code):
    exec('global i; i = %s' % code)
    global i
    return i

This is misusing global variables to get your data across the border.

>>> my_exec('1 + 2')
3

Needless to say that you should never allow any user inputs for the input of this function in there, as it poses an extreme security risk.

Solution 4

Something like this can work:

def outer():
    def inner(i):
        return i + 10


for f in outer.func_code.co_consts:
    if getattr(f, 'co_name', None) == 'inner':

        inner = type(outer)(f, globals())

        # can also use `types` module for readability:
        # inner = types.FunctionType(f, globals())

        print inner(42) # 52

The idea is to extract the code object from the inner function and create a new function based on it.

Additional work is required when an inner function can contain free variables. You'll have to extract them as well and pass to the function constructor in the last argument (closure).

Solution 5

Here's a way to return a value from exec'd code:

def exec_and_return(expression):
    exec(f"""locals()['temp'] = {expression}""")
    return locals()['temp']

I'd advise you to give an example of the problem you're trying to solve. Because I would only ever use this as a last resort.

View more solutions

Comments

[prashant]

I don't think you can do this with exec. You'll need to do something like georg's answer or use new as described in this answer.

[gog]

@DavidMoles: new is deprecated, but types are fine, thanks.

[marsh]

That is so weird, Especially that you dont need to declare the variable before hand.

[Chris]

This isn't working for me. exec('ds_id = kim_processors.%s.process(%s)' % (function.python_module, model_id)) does populate ds_id, as I can see in the Django locals debug output. However I then run cursor.execute("""select e.id as ... where dataset_id=%s""", (ds_id)) and get error "Exception Type: NameError Exception Value: name 'ds_id' is not defined". Yet locals shows ds_id is indeed set (to 14). Is there some restriction on its scope?

[Chris]

Also tried exec('import kim_processors.%s as processor_module' % function.python_module) then running ds_id = processor_module.process(model_id). Again, I can see processor_module listed in localvars, but I get the same NameError exception "Exception Value: name 'processor_module' is not defined"

[wnnmaw]

@Chris, I'm not sure what the issue is, but it sounds like a good question for the site. Try posting it separately as a question and refer here to show what you've tried.

[Chris]

Thanks @wnnmaw that's exactly what I did! Sorry forgot to follow up on this thread. The question and answers are here There is a better way of importing (importlib) and the duplicate question describes the scope issue beautifully. Thanks for responding.

[Chris Fowl]

Thank you! +1. I struggled a lot with getting a reference to a module given a string and this is what works. You can use this by giving the string as parameter code and it will return the actual module.

[misantroop]

This doesn't work, exec needs to be launched with globals, exec("a = foo()", globals())

[wnnmaw]

@misantroop, just tested on Python 3.7.0 and 3.8.1, works fine on both. What system are you using where it doesn't work?

[Doug]

@wnnmaw this doesn't work in general only when a is a global variable. You can see it not working by simply moving your code into a def foo(): ... exec('a ==' statement. May I suggest updating your answer to show that it only works when the variable you modify is a global? eg. In a function you need to, eg. def foo(): exec('global a\na = 5').

[GoodJuJu]

Whilst your answer may be useful, the question was asked 6 years ago and already has 5 answers, although none have been accepted.

[Patrick McGloin]

This is good. But you could just use this code and not have a seperate dictionary for "loc": exec(the_code) return_workaround = locals()['return_me']

[Ank]

Similar answers are already there posted. Better to look at new questions to answer where community needs you.

[Namwon Kim]

@Ank You are so right. But, I also want to share a simple method. Thank you.

[bomben]

This works, but inside the code I am unable to reference to variables that where defined outside it.

[Mr. B.]

@bomben I didn't test it, but can't you pass your outer variables? It should be mentioned in the docs of exec(). :)

[Mathieu Longtin]

or: return eval(expression)

[Mr. B.]

@aceminer Happy to read that! :)