rgolwalkar rgolwalkar - 4 months ago 27
Python Question

Explain Keyword-Only Arguments (VarArgs) in Python

Please see the code below:-

# Filename: total.py

def total(initial=5, *numbers, **keywords):
count = initial
for number in numbers:
count += number
for key in keywords:
count += keywords[key]
return count

print(total(10, 1, 2, 3, vegetables=50, fruits=100))

Can someone please explain how is *numbers and **keywords picking up the arguments?
A simple explaination is very much appreciayed
Thanks in advance


In your code numbers is assigned to the (1,2,3) tuple. keywords is assigned to a dictionary, containing vegetables and fruits.

One star '*' defines positional arguments. This means that you can receive any number of arguments. You can treat the passed arguments as a tuple.

Two stars '**' define keywords arguments.

The reference material is available here.


Python 2.x (before keyword-only arguments)

def foo(x, y, foo=None, *args): print [x, y, foo, args]

foo(1, 2, 3, 4)            --> [1, 2, 3, (4, )]  # foo == 4
foo(1, 2, 3, 4, foo=True)  --> TypeError

Python 3.x (with keyword-only arguments)

def foo(x, y, *args, foo=None): print([x, y, foo, args])

foo(1, 2, 3, 4)           --> [1, 2, None, (3, 4)]  # foo is None
foo(1, 2, 3, 4, foo=True) --> [1, 2, True, (3, 4)]

def combo(x=None, *args, y=None): ...  # 2.x and 3.x styles in one function

Although a seasoned programmer understands what happened in 2.x, it's counter-intuitive (a positional argument gets bound to foo= regardless of keyword arguments as long as there are enough positional arguments)

Python 3.x introduces more intuitive keyword-only arguments with PEP-3102 (keyword arguments after varargs can only be bound by name)