Function call and the call attribute

Question

Python's callables are objects that have a __call__ method. They are most of the time functions, but might be class instances as well.

But it so happens that functions do have a __call__ method. Therefore, a __call__ method has a __call__ method as well. The following REPL session shows that we can chain the __call__s:

>>> print
<built-in function print>

>>> print.__call__
<method-wrapper '__call__' of builtin_function_or_method object at 0x0000025E2D597F78>

>>> print.__call__.__call__
<method-wrapper '__call__' of method-wrapper object at 0x0000025E2F631438>

>>> print.__call__.__call__.__call__
<method-wrapper '__call__' of method-wrapper object at 0x0000025E2F5F85F8>

>>> print.__call__.__call__.__call__.__call__
<method-wrapper '__call__' of method-wrapper object at 0x0000025E2F725DA0>

... and so on. Remarkably, all these methods have different addresses. In addition, they all have the same behaviour:

>>> print("a")
a
>>> print.__call__("a")
a
>>> print.__call__.__call__("a")
a
>>> print.__call__.__call__.__call__("a")

So, when I write print("a"), what is actually called? Is it print, or print.__call__? And what if I define a Foo class with a __call__ method?

Furthermore, how is it possible that every __call__ method has its own different __call__ method? Could it be that they are in fact created when I tried to access them?

They have different addresses because they are descriptors, and are bound to the function object instance when accessed. — Martijn Pieters, Sep 11 '17 at 16:15
[Close enough?](https://stackoverflow.com/q/32855927/846892) — Ashwini Chaudhary, Sep 11 '17 at 16:17

score 3 · Accepted Answer · edited Sep 11 '17 at 16:24

Like methods on classes, the __call__ attribute is a descriptor object defined on the type, bound to the object you looked it up on:

>>> type(print).__dict__['__call__']
<slot wrapper '__call__' of 'builtin_function_or_method' objects>
>>> type(print).__dict__['__call__'].__get__(print)
<method-wrapper '__call__' of builtin_function_or_method object at 0x10cc66f78>
>>> print.__call__
<method-wrapper '__call__' of builtin_function_or_method object at 0x10cc66f78>

The binding behaviour (through the __get__ method) is how the resulting method-wrapper instance knows to pass in the print object as self, just like the instance gets passed into methods you defined on a custom Python class.

So yes, that means they are created on demand, and thus will have a unique id. They are otherwise just more instances of the same type.

score 3 · Answer 2 · answered Sep 11 '17 at 16:18

print is an instance of the class builtin_function_or_method:

>>> type(print)
builtin_function_or_method

As we all know (hopefully) when you access a method the instance is implicitly passed as first argument:

>>> type(print).__call__(print, 10)
10
>>> print.__call__(10)  # equivalent
10

That means you actually have something like a bound method when you access __call__ of a function. However bound methods (or in this case methodwrappers) are also instances of the type method_wrapper:

>>> type(print.__call__)
method-wrapper

So when you access the __call__ method of that one it's again a "new" bound method.

>>> type(print.__call__).__call__(print.__call__, 10)  # equivalent
10

At this point it becomes recursive because a method of a method-wrapper is just another method-wrapper instance:

>>> type(print.__call__.__call__)
method-wrapper

However all that does it just creates a lot of "unnecessary" bound methods...

Function call and the __call__ attribute

2 Answers2

Function call and the call attribute