Unexpected behaviour with decorator

Question

Let's suppose I have a simple function returning its own name

def foo():
    print(foo.__name__)

Of course the output when calling is foo. However, if we now decorate that function with a decorator like this:

def dec(func):
    print(dec.__name__)

    def wrapper(*args):
        print(wrapper.__name__)
        return func(*args)
    return wrapper


@dec
def foo():
    print(foo.__name__)


foo()

We get

dec
wrapper
wrapper

Why is the name of the inner function called here? func(*args) is specifically called, shouldn't it be really the function's name among with dec wrapper?

Does this answer your question? [Get the name of a decorated function?](https://stackoverflow.com/questions/4887081/get-the-name-of-a-decorated-function) — wjandrea, May 29 '21 at 22:42
I would like to know the reason, not a solution to this occurence — Raumschifffan, May 29 '21 at 22:44
Oh, I see. Does this answer your question? [What does functools.wraps do?](https://stackoverflow.com/q/308999/4518341) *"When you use a decorator, you're replacing one function with another."* — wjandrea, May 29 '21 at 22:47
Do you understand what decorators do? You've assigned the wrapper function to `foo` so *of course* `foo.__name__` is wrapper — juanpa.arrivillaga, May 29 '21 at 23:33
@juanpa.arrivillaga Yes, now I do. I just always set a decorator equal with (in this example): `dec(foo)()` — Raumschifffan, May 29 '21 at 23:44

score 1 · Answer 1 · answered May 29 '21 at 22:48

As for the reason what the decorator does is replace your original function with the wrapper. So in fact foo isn’t foo anymore, it is the closure created by combining wrapper and func.

wraps is a trick to transfer such metadata to the newly created function.

Unexpected behaviour with decorator

1 Answers1