How to pass decorator default value to inner function in Python

Question

I’m learning about function decorators and am trying to apply some of the concepts I’ve learned so far. For the below function i’ve created a decorator function but it doesnt behave as I would expect.

Can someone help me understand why I can’t access the default value (or any provided arg) of the outer function so that I can use it in the inner function? I keep getting a TypeError: type function doesn't define __round__ method.

def mathTool(defaultVal=5):
        def innerFunc(*args):
            return round(*args, defaultVal)
        return innerFunc

print(funcB()) //triggers TypeError
print(funcB(2)) //triggers TypeError

def funcA(A):
    return A/2

@mathTool()
def funcB(B):
    return B/2

def funcAB(A, B):
     return A*B/2

func_dec = mathTool(defaultVal=7)(funcAB)

print(func_dec(5, 9)) //triggers TypeError

Asked By: SwingingDonkey

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Source

Answer 1

Decorators take exactly one argument: the function. And their syntax is this:

@deco
def func():
    ...

When you see a decorator which takes arguments, as in @deco(arg), that is actually a function which returns a decorator.

Therefore:

def mathTool(defaultVal=5):
    def decorator(func):
        def innerFunc(*args):
            args = args + (defaultVal,)
            return func(*args)
        return innerFunc
    return decorator

@mathTool(7)
def whatever(a, b):
    ...

Now, calling whatever(1) will pass 1 and 7 to the function.

Answered By: zvone

How to pass decorator default value to inner function in Python

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