How do I use method overloading in Python

0 votes

I am trying to implement method overloading in Python:

class A:
    def stackoverflow(self):    
        print 'first method'
    def stackoverflow(self, i):
        print 'second method', i

ob=A()
ob.stackoverflow(2)

but the output is second method 2; similarly:

class A:
    def stackoverflow(self):    
        print 'first method'
    def stackoverflow(self, i):
        print 'second method', i

ob=A()
ob.stackoverflow()

gives

Traceback (most recent call last):
  File "my.py", line 9, in <module>
    ob.stackoverflow()
TypeError: stackoverflow() takes exactly 2 arguments (1 given)

How do I make this work?

Jan 3, 2023 in Python by erzan
• 630 points
272 views

1 answer to this question.

0 votes

It's method overloading, not method overriding. And in Python, you historically do it all in one function:

class A:
    def stackoverflow(self, i='some_default_value'):
        print('only method')

ob=A()
ob.stackoverflow(2)
ob.stackoverflow()

See the Default Argument Values section of the Python tutorial. See "Least Astonishment" and the Mutable Default Argument for a common mistake to avoid.

See PEP 443 for information about the single dispatch generic functions added in Python 3.4:

>>> from functools import singledispatch
>>> @singledispatch
... def fun(arg, verbose=False):
...     if verbose:
...         print("Let me just say,", end=" ")
...     print(arg)
>>> @fun.register(int)
... def _(arg, verbose=False):
...     if verbose:
...         print("Strength in numbers, eh?", end=" ")
...     print(arg)
...
>>> @fun.register(list)
... def _(arg, verbose=False):
...     if verbose:
...         print("Enumerate this:")
...     for i, elem in enumerate(arg):
...         print(i, elem)

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answered Jan 4, 2023 by Elton
• 400 points

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