why fct.__call__() is not the same as fct()?-CodePudding

the __call__ dunder method of a class is supposed to represent the operator (), correct? if so, why the following code is not working as expected?


def f():
    print("hello world")

f()
>>>hello world

f.__call__ = lambda: print("foo") #replace the call operator

f()
>>>hello world

f.__call__()
>>>foo

something special is happening with <class 'function'> could someone shine a light on this problem?

CodePudding user response：

You have not changed the __call__ method of a class. You have changed the __call__ method of a class instance. That's the difference. You would need to change the __call__ method of the function class, which of course you don't really want to do.

The documentation on this is terse:

    Class Instances
        Instances of arbitrary classes can be made callable by defining
        a __call__() method in their class.

It specifically says "in their class", and not "in the instance". This is just a design decision on Python's part. You can imagine how confusing it would be if different instances of a class behaved differently in that case.