CODEWITHSUNDEEP
pythonpython-3.xparameterspython-decorators
LaN tHe MaN
LaN tHe MaN

Reputation: 45

Do we have to use the function value in wrapped function of a decorator?

def div2(mult):
    def display():

        print(f"we are doing a check of multiples if div by 2")
        if mult%2==0:
            return f"{mult} is divisible"
    return display

@div2
def multi(n):
    return n*n*n

print(multi(5))

In a video i learnt that we need not pass an argument for the wrapped function(display in this e.g), the main function's argument(div2 in this case) will be used in the wrapped function. When i tried the same for the above code i got a TypeError that display() needs a parameter.

Confused now that in this case should all my 3 functions should have the parameters as below: div2() -> multi() function itself as parameter display() -> a parameter to display the value of function multi() multi() -> a parameter to do the math

Note: When I made display() take a parameter, the code worked

Upvotes: 0

Views: 38

Answers (2)

ruohola
ruohola

Reputation: 24058

Here multi takes one parameter, named n.

def multi(n):
    return n*n*n

And using the decorator syntax is actually the same as doing:

multi = div2(multi)

Thus if you want be able to pass paramaters to the wrapper display it has to be defined to take the parameters.

Upvotes: 0

chepner
chepner

Reputation: 531420

The problem is not that display doesn't take an argument; it's that you use it as if it does. Remember that decorator syntax is just syntactic sugar for

def multi(n):
    return n*n*n

multi = div2(multi)

The result of div2 isn't required to preserve the signature of div2's argument, though you have to know if it does or not before you can use that result.

The argument to div2 is a function, not something you can divide by 2. Your decorator should be

def div2(f):
    def display(x):
        print(f"We are doing a check of multiples if div by 2")
        if f(x) % 2 == 0:
            return f"{x} is divisible"
    return display

@div2
def multi(n):
    return n*n*n

prin(multi(5))

Upvotes: 1

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