How to specify in a python function that a passed argument must be a function

Question

I've made a function that takes a function as an argument.

If I want to specify that a function takes an integer it's pretty easy:

def fun(myInt:int=0):
    return myInt

But when it takes a function I can't do this:

def fun(myFun:function=print):
    myFun('whatever')

I printed type() of both an integer and a function and got<class 'int'> <class 'function'>

What is the difference between these two and is an easy way to do what I wanted to do?

Btw. I've already used fellow google and couldn't find anything

You are using *type hinting* and the correct spelling would be to use a [`typing.Callable`](https://docs.python.org/3/library/typing.html#typing.Callable) generic. — Martijn Pieters, Jan 11 '19 at 15:48
Alrighty then, thanks a lot! (Looks like what I needed was to know the name of what I was looking for) — Ян Никодим Лудян Печул, Jan 11 '19 at 19:22

score 1 · Accepted Answer · answered Jan 11 '19 at 15:58

1

In Python 3:

from typing import Callable

def fun(myFun:Callable=print):
    myFun('whatever')

fun()
fun(print)
fun(type)

answered Jan 11 '19 at 15:58

1 Answers1