Why function name returns wrapper function name and not the name of the function the code sits in?

Question

def logging_decorator(function):
    def wrapper(*args, **kwargs):
        print(f'This is {function.__name__}')
        output = function(*args, **kwargs)
    return wrapper


@logging_decorator
def a_function(a, b, c):
    print(f'This is the Function name {a_function.__name__}')
    return a * b * c

object = a_function(3, 4, 5)

Been playing around with python decorators and I don't understand the following:

Why does

print(f'This is the Function name {a_function.__name__}')

output 'This is the Function name wrapper'? Shouldn't the output be 'This is the Function name a_function' instead?

No, obviously not. What function do you think you are calling? — juanpa.arrivillaga, Sep 28 '21 at 20:33
The name has to be of the wrapper; it’s wrapping the decorated function — DisappointedByUnaccountableMod, Sep 28 '21 at 20:34
The `@deco def foo` syntax is equivalent to ` def foo()... foo = deco(foo)`, what does your decorator return? — juanpa.arrivillaga, Sep 28 '21 at 20:35
I'm calling the a_function.__name__ i'm not calling the wrapper.__name__ yet it still prints the wrapper function name — Paul Ged, Sep 28 '21 at 20:36
Try answers https://stackoverflow.com/questions/147816/preserving-signatures-of-decorated-functions — DisappointedByUnaccountableMod, Sep 28 '21 at 20:36
The variable `a_function` *refers to the value returned from your decorator*, which is the function you define inside as `def wrapper` and then you `return wrapper`, this is **crucial** to understand — juanpa.arrivillaga, Sep 28 '21 at 20:38

Why function __name__ returns wrapper function name and not the name of the function the code sits in?

0 Answers0

Why function name returns wrapper function name and not the name of the function the code sits in?