Python functions and their call attribute

Question

I am using Python 2.7.2. I want to understand the relationship between calling a function and calling the __call__ attribute of the function. For example, consider the following code

def foo():
    return 5
print foo()          # ==> 5
print foo.__call__() # ==> 5

foo.__call__ = lambda : 6
print foo()          # ==> 5
print foo.__call__() # ==> 6

The firsts four lines seem to indicate the calling the function foo is the same as calling the __call__ attribute of foo. However, the last three lines seem to indicate that they are different beasts since I changed the __call__ attribute but it didn't change the value returned by a call to foo().

Can someone explain the relationship between calling foo() and calling foo.__call__()? Is there a way to modify the behavior of the function so that foo() as well as foo.__call__() now returns 6 instead of 5?

There's never a reason to modify foo.__call__ -- you could just modify foo instead! — Katriel

gcbirzan gcbirzan · Accepted Answer · 2012-03-14T18:10:15

I recommend you read special method lookup for new-style classes (especially the last paragraph).

For new-style classes, implicit invocations of special methods are only guaranteed to work correctly if defined on an object’s type, not in the object’s instance dictionary.

Python functions and their __call__ attribute

1 Answers

Python functions and their call attribute