CODEWITHSUNDEEP
pythonpython-3.xfunctionnested
invest eatstudy
invest eatstudy

Reputation: 11

Python nested function is not equal to itself

This simple Python code gets "False".

def foo():
    def bar():
        return 0
    return bar
print(foo() == foo())

When I request 

print(foo(),foo())

I get

<function foo.<locals>.bar at 0x03A0BC40> <function foo.<locals>.bar at 0x03C850B8>

So does Python store the result of the bar function every time in the new memory slot? I'd be happy if someone explain how it works behind the scene and possibly how this code can be a little bit modified to get "True" (which still seems logical to me!).

Upvotes: 1

Views: 52

Answers (2)

RafalS
RafalS

Reputation: 6354

Each def statement defines a new function. The same name and body doesn't really matter. You're basically doing something like this:

def foo():
    pass

old_foo = foo

def foo():
    pass

assert old_foo == foo # will fail

Upvotes: 1

Mureinik
Mureinik

Reputation: 312404

bar is locally defined within foo the same way a local variable would be each time you call the foo().

As you can see by the defult __repr__, they have different memory addresses, and since bar does not implement __eq__, the two instances of bar will not be equal.

Upvotes: 0

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