103

Is it possible to have overloaded functions in Python?

In C# I would do something like

void myfunction (int first, string second)
{
    # Some code
}

void myfunction (int first, string second, float third)
{
    # Some different code
}

And then when I call the function it would differentiate between the two based on the number of arguments. Is it possible to do something similar in Python?

Peter Mortensen
  • 30,738
  • 21
  • 105
  • 131
Trcx
  • 4,164
  • 6
  • 30
  • 30
  • possible duplicate of [Function overloading in Python: Missing](http://stackoverflow.com/questions/733264/function-overloading-in-python-missing) – Li0liQ Aug 18 '11 at 19:35
  • This seems to be a possible duplicate post. Also please don't flag as C# as the question doesn't have to do with C#. [function-overloading-in-python-missing](http://stackoverflow.com/questions/733264/function-overloading-in-python-missing) – Jethro Aug 18 '11 at 19:34
  • possible duplicate of [Python function overloading](http://stackoverflow.com/questions/6434482/python-function-overloading) – Ciro Santilli OurBigBook.com Jul 14 '15 at 18:48
  • Related (not duplicate): *[How can I detect duplicate method names in a Python class?](https://stackoverflow.com/questions/10761988)* – Peter Mortensen Jan 06 '21 at 16:20

6 Answers6

116

EDIT For the new single dispatch generic functions in Python 3.4, see http://www.python.org/dev/peps/pep-0443/

You generally don't need to overload functions in Python. Python is dynamically typed, and supports optional arguments to functions.

def myfunction(first, second, third = None):
    if third is None:
        #just use first and second
    else:
        #use all three

myfunction(1, 2) # third will be None, so enter the 'if' clause
myfunction(3, 4, 5) # third isn't None, it's 5, so enter the 'else' clause
agf
  • 171,228
  • 44
  • 289
  • 238
  • 4
    But note that calling different functions based on the *type* of the arguments is much more difficult (although not impossible). – Andrew Jaffe Sep 05 '11 at 13:37
  • 11
    Well, there is a difference between overloading/polymorphism and conditionals. Do you mean that we do not need the polymorphysm since we have the conditionals? – Val Oct 15 '13 at 11:02
  • 1
    @Val I'm saying, basically, that in a dynamically typed language you don't need overloading as a language feature because you can trivially emulate it in code, and so can get polymorphism that way. – agf Oct 15 '13 at 16:04
  • I do not see how dynamic linking answers my question. Now, since java is statically typed, you say that in cannot have the optional arguments. I do not understand this also. – Val Oct 15 '13 at 16:58
  • 1
    @Val I'm saying that between optional arguments and dynamic types the way you have them in Python, you don't need Java style method overloading to achieve polymorphism. – agf Oct 15 '13 at 18:30
  • What if I want one version of the method to return a generator, and the other version "return" a single object, depending on the parameters? Yielding and returning from the same method is not permitted in python 2.x. – navidoo Nov 21 '14 at 03:13
  • 3
    @navidoo I would argue that isn't a good idea -- a function shoudn't return completely different types of things. However, you can always define a local generator function inside the main function, call it, and return the resulting generator -- from the outside it will be the same as if the main function was a generator. [Example here.](https://gist.github.com/agfor/a81e60d5ae4bd8db70ca) – agf Nov 21 '14 at 04:45
  • 1
    @agf What about the case where you need def searchItem(self, item_key) and searchItem(self, item_description)? – goul May 25 '17 at 08:54
  • @goul Generally, this is what keyword arguments are for. `def search_item(self, item_key = None, item_description = None)`. On Python 3, you can make them keyword only to avoid the confusion of them being specified positionally -- `def search_item(self, *, item_key = None, item_description = None)`. There are also many other ways to solve this type of problem, like having two subclasses to handle the different cases, just two different methods, etc. It depends on the structure of the rest of the code. – agf May 26 '17 at 04:30
58

In normal Python you can't do what you want. There are two close approximations:

def myfunction(first, second, *args):
    # 'args' is a tuple of extra arguments

def myfunction(first, second, third=None):
    # 'third' is optional

However, if you really want to do this, you can certainly make it work (at the risk of offending the traditionalists ;o). In short, you would write a wrapper(*args) function that checks the number of arguments and delegates as appropriate. This kind of "hack" is usually done via decorators. In this case, you could achieve something like:

from typing import overload

@overload
def myfunction(first):
    ....

@myfunction.overload
def myfunction(first, second):
    ....

@myfunction.overload
def myfunction(first, second, third):
    ....

And you'd implement this by making the overload(first_fn) function (or constructor) return a callable object where the __call__(*args) method does the delegation explained above and the overload(another_fn) method adds extra functions that can be delegated to.

You can see an example of something similar here http://acooke.org/pytyp/pytyp.spec.dispatch.html, but that is overloading methods by type. It's a very similar approach...

And something similar (using argument types) is being added to Python 3 - PEP 443 -- Single-dispatch generic functions

Peter Mortensen
  • 30,738
  • 21
  • 105
  • 131
andrew cooke
  • 45,717
  • 10
  • 93
  • 143
  • 1
    The `range` builtin uses overload, so is it really a hack? https://github.com/python/typeshed/blob/master/stdlib/2and3/builtins.pyi#L1051 – CptJero May 18 '19 at 00:10
  • Is this a mistake: `@myfunction.overload`? Should not just all of them have `@overload`, and you would then need another function without the `@overload` on top of this? I have just tested `@overload` in quite a few settings. I wonder whether this answer is outdated. `typing` does not allow you to overload in the code, it is just to show what the type hints would give you *if* you overloaded the function. Which you do not, since `typing` is nothing about the working code, instead, it is about making code quickly readable. – questionto42 May 22 '23 at 00:15
  • I dared to answer at [How can I type-hint a function where the return type depends on the input type of an argument?](https://stackoverflow.com/a/76302414/11154841). Correct me if I am wrong, but `typing`, in my humble opinion, does not change the working of the code, it is just to quickly show what is going on. There is no overloading with `typing`, the `typing` overload functions are not run at all, as far as I can see. I could be wrong. – questionto42 May 22 '23 at 00:25
11

Yes, it's possible. I wrote the code below in Python 3.2.1:

def overload(*functions):
    return lambda *args, **kwargs: functions[len(args)](*args, **kwargs)

Usage:

myfunction=overload(no_arg_func, one_arg_func, two_arg_func)

Note that the lambda returned by the overload functions choose a function to call depending on the number of unnamed arguments.

The solution isn't perfect, but at the moment I can't write anything better.

Peter Mortensen
  • 30,738
  • 21
  • 105
  • 131
GingerPlusPlus
  • 5,336
  • 1
  • 29
  • 52
  • Great way to refresh the args and kwargs thingy... Yet, is it pythonic to make it to everyday code? I doubt it. I guess in Python, they prefer `try` and `except` or `if` and `else` or optional (`=None`) over this small overload overhead, so that your good code was avoided from the start of Python. – questionto42 May 22 '23 at 09:43
1

It is not possible directly. You can use explicit type checks on the arguments given though, although this is generally frowned upon.

Python is dynamic. If you are unsure what an object can do, just try: and call a method on it, then except: errors.

If you don't need to overload based on types, but just on the number of arguments, use keyword arguments.

Peter Mortensen
  • 30,738
  • 21
  • 105
  • 131
Jürgen Strobel
  • 2,200
  • 18
  • 30
1

Here is the way to overload python functions with default arguments as well as keyword arguments

from multipledispatch import dispatch

# FOR hi(a: int, b: int = 3)

@dispatch(int, int)
def _hi(a: int, b: int):
    print(a, b)


@dispatch(int, int)
def hi(a: int, b: int = 3):
    _hi(a, b)


@dispatch(int, b=int)
def hi(a: int, *, b: int = 3):
    _hi(a, b)


# FOR hi(a: str, b: int = 3)

@dispatch(str, int)
def _hi(a: str, b: int):
    print(a, b, 'str!')


@dispatch(str, int)
def hi(a: str, b: int = 3):
    _hi(a, b)


@dispatch(str, b=int)
def hi(a: str, *, b: int = 3):
    _hi(a, b)

hi(2)
hi(2, 3)
hi(2, b=3)
hi('2')
hi('2', 3)
hi('2', b=3)

Output

2 3
2 3
2 3
2 3 str!
2 3 str!
2 3 str!
0

Overloading methods is tricky in Python. However, there could be usage of passing the dict, list or primitive variables.

I have tried something for my use cases, and this could help here to understand people to overload the methods.

Let's take the example use in one of the Stack Overflow questions:

A class overload method with call the methods from different class.

def add_bullet(sprite=None, start=None, headto=None, spead=None, acceleration=None):

Pass the arguments from a remote class:

add_bullet(sprite = 'test', start=Yes, headto={'lat':10.6666, 'long':10.6666}, accelaration=10.6}

Or

add_bullet(sprite = 'test', start=Yes, headto={'lat':10.6666, 'long':10.6666}, speed=['10','20,'30']}

So, handling is being achieved for list, Dictionary or primitive variables from method overloading.

Try it out for your code.

Peter Mortensen
  • 30,738
  • 21
  • 105
  • 131
shashankS
  • 1,043
  • 1
  • 11
  • 21