Passing instance method as default argument

Question

Maybe this is a bit funky, but I want to pass a callable instance method as a default variable to an instance method of the same class.

from typing import Callable

class Person:

    def __init__(self, name):
        self.name = name

    def say_hi(self):
        print("Hello, my name is", self.name)

    def greet(self, func: Callable=say_hi):
        func()

p = Person("Guido")
p.greet()

With the above solution self is not passed to the instance method. Replacing func: Callable=say_hi with func: Callable=self.say_hi and self is not yet defined.

if you know it is an instance method? why are you passing it? You can access it inside greet like self.say_hi, if you still insist then you can d0 p.greet(p.say_hi) — Epsi95
– Epsi95, Commented Sep 10, 2021 at 8:46

Matiiss · Accepted Answer · 2021-09-10 10:04:55Z

3

You can do assignment in the function itself:

def greet(self, func: Callable = None):
    func = self.say_hi if func is None else func
    func()

Suggestion by @blhsing in the comments below is to use this (tho being explicit is kinda better than implicit, but this will work too):

func = func or self.say_hi

edited Sep 10, 2021 at 10:04

answered Sep 10, 2021 at 8:49

Matiiss

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1 Comment

blhsing Over a year ago

Or func = func or self.say_hi if you like it shorter. (Too bad there's no equivalent of the ||= operator of Ruby in Python.)

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