Cunfusion with method overloading in python?

Question

I am new to python. But I know about method overloading. But I am confusion with overloading in python. Here my confusion code,

class OverLoad(object):
    """docstring for OverLoad"""
    def __init__(self, arg):
        super(OverLoad, self).__init__()
        self.arg = arg

    def adder(a, b):
        print a,b

    def adder(*a):
        print a

    def adder(a):

        print a
    def adder():

        print "no arg"

And please explain the above code.

Python doesn't have method overloading - the last method you define is the one that's used... that could be why? :) Maybe have a look at: http://stackoverflow.com/questions/6434482/python-function-overloading — Jon Clements, Dec 05 '13 at 08:36

sheh · Answer 1 · 2013-12-05T08:45:47.403

1

Python-style overloading:

def adder(self, *arg, **kwd):

you can call:

some_class.adder(1, 2)
some_class.adder(1, 2, 3, 4, 5, 6, 7, 9, 10 ....)
some_class.adder(1, 2, 3, 4, arg1=5, arg2=6, arg3=7)
some_class.adder(arg1=1, arg2=2, arg3=3)

but most likely your variant:

def adder(self, *arg):
    if len(arg) == 0:
       print "no arg"
    return sum(arg)

and call:

some_class.adder(1,2,3,4,5,7,8,9,10)

edited Dec 05 '13 at 08:45

answered Dec 05 '13 at 08:39

sheh

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score 0 · Answer 2 · answered Dec 05 '13 at 08:50

Python does not support method overloading.

Refer to this doc. When you define a method in a class, Python creates a function object first and then binds the function object and the function name that you defines. So in your code example, only the last adder() function takes effect.

Cunfusion with method overloading in python?

2 Answers2