Decorator with arguments

Question

def Decorator(*pos, **kwargs):
    def innerFunc(arg, num):
        arg()
        print('Inner function')
        print(kwargs['name'])
    return innerFunc

@Decorator(name = 'Michael')
def inputFunc():
    print('Input Function')

Why does applying the Decorator to inputFunc automatically call it? I would like to pass in some arguments to the decorated function but cannot because the Decorator already calls it

That's just the way Python decorators work, if you want to pass arguments, you need to add a 2nd wrapper between `inner_func` and `decorator`, call it `outer_func`. Google around and you'll find it. — Mark Mikofski, Feb 08 '20 at 22:55
`@decorator def f(): ...` is equivalent to `def f(): ...` and then `f = decorator(f)`, so that's why the `decorator` function is called immediately. — kaya3, Feb 08 '20 at 22:58
It doesnt call the innerFunc when i dont pass in arguments to the decorator — Streamline Astra, Feb 08 '20 at 23:02

score -3 · Answer 1 · answered Feb 08 '20 at 22:58

-3

In your example, remove the arg() call:

def Decorator(*pos, **kwargs):
    def innerFunc(arg, num):
        # arg()  <-- this is calling the function
        print('Inner function')
        print(kwargs['name'])
    return innerFunc

@Decorator(name = 'Michael')
def inputFunc():
    print('Input Function')

answered Feb 08 '20 at 22:58

Tiger-222

6,677
3
47
60

1

This isn't really right; if you run the OP's code, it gives a `TypeError` on calling `innerFunc` so `arg` never actually gets called. This also isn't a correct way to write a decorator which takes arguments, and if you don't use `arg` then it's not really a proper decorator at all. You need two levels of nested functions, one returning a decorator, which returns a wrapped function. – kaya3 Feb 08 '20 at 23:00
I answered to the "Why does applying the Decorator to inputFunc automatically call it?" question. You are right, this is not a good example nor the good way to create a decorator. – Tiger-222 Feb 08 '20 at 23:13
1

But your answer doesn't answer it; the OP's code never actually calls `arg` because there is a `TypeError` first. Your code raises the same `TypeError` for the same reason. The OP's code calls `innerFunc`, and your answer doesn't explain why. – kaya3 Feb 08 '20 at 23:14

Decorator with arguments

1 Answers1