function 作为可选参数 python

Question

i am trying to write a function to check if a parameter was passed to it (which is a function ) if so call that function with an argument else if there wasn't any argument given return a value so my approach was like this:

def nine(fanc=None):
    if(fanc!=None): return fanc(9,fanc) 
    return 9

but this code rise an error which is:

TypeError: 'int' object is not callable

i know that this approach isn't correct but i couldn't find any other way to do so i have also tried using *args this way but end up with the same results:

def nine(*args):
    if(len(args)!=0): return args[0](9,args) 
    return 9

Answer 1

I try to guess what you want but this snippet might help you:

def fct(**kwargs):
    if 'func' in kwargs:
        f = kwargs['func']
        return f(9)
    else:
        return 9

def f(x):
    return x**2
    
print(fct())  # result = 9
print(fct(func=f))  # result = 81

Answer 2

You might use callable built-in function, consider following example

def investigate(f=None):
    if callable(f):
        return "I got callable" 
    return "I got something else"
print(investigate())
print(investigate(min))

output:

I got something else
I got callable

Beware that callable is more broad term that function, as it also encompass objects which have __call__ method.

Answer 3

If you want to check whether the passed argument is a function and, if yes, then execute it with fixed arguments, you could try the following:

from typing import Union
from types import FunctionType

def nine(func: Union[FunctionType, None] = None):
    if type(func) is FunctionType:
        return func(9) 
    return 9