Problem while defining a function in a `for` loop in Python

Question

I have the following minimal example that reproduces the issue:

def f1(a, b):
    print(1)
    return a + b


def f2(a, b):
    print(2)
    return a * b


funcs = f1, f2


_funcs = []
for func in funcs:

    def _func(x):
        return func(*x)

    _func.__name__ = func.__name__
    _funcs.append(_func)
funcs = _funcs


for func in funcs:
    print(func([2, 3]))

This throws a TypeErro:

TypeError: _func() takes 1 positional argument but 2 were given

with the following traceback:

---------------------------------------------------------------------------

TypeError                                 Traceback (most recent call last)

<ipython-input-93-e24e67b5e525> in <module>()
     22 
     23 for func in funcs:
---> 24     print(func([2, 3]))

<ipython-input-93-e24e67b5e525> in _func(x)
     14 
     15     def _func(x):
---> 16         return func(*x)
     17 
     18     _func.__name__ = func.__name__

TypeError: _func() takes 1 positional argument but 2 were given

I am not sure why as it seems to me that _func() is always called with a single argument.

I suspect what I want to do would be done with functools, but still I fail to see why the above code works this way. Any hint?

Very strange, and when you type print(funcs[0]([2, 4])) and print(funcs[1]([2, 4])) (so outside the for loop) in both case, I get 2 and 8. — Léonard
– Léonard, Commented Sep 30, 2020 at 20:59

juanpa.arrivillaga · Accepted Answer · 2020-09-30 21:26:55Z

3

You are closing over the free-variable func when you define:

def _func(x):
    return func(*x)

However, in your loop, you re-using this name,

for func in funcs:
    print(func([2, 3]))

So func now refers to the single-argument functions you defined, even inside _func! So of course, when that function calls func(*x) it's actually recursively calling itself. Use a different name:

for f in funcs:
    print(f([2, 3]))

Or better yet, don't rely on a free-variable you don't intend to be free:

def func_maker(func):
    def _func(x):
        return func(*x)
    return _func

And use that:

...

_funcs = []
for func in funcs:

    _func = func_maker(func)

    _func.__name__ = func.__name__
    _funcs.append(_func)
funcs = _funcs

...

edited Sep 30, 2020 at 21:26

answered Sep 30, 2020 at 20:51

juanpa.arrivillaga

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2 Comments

0x5453 Over a year ago

See also: Why do lambdas defined in a loop with different values all return the same result?

juanpa.arrivillaga Over a year ago

@0x5453 yes, it's the same fundamental issue, although, I think in this case it isn't as obvious. I had to look it over a couple of times.

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