[Tutor] Defining variable arguments in a function in python

[Peter Otten]

Karthik Bhat wrote:

> Hello,
> 
>         I have the following piece of code. In this, I wanted to make use
> of the optional parameter given to 'a', i.e- '5', and not '1'
> 
> def fun_varargs(a=5, *numbers, **dict):
>     print("Value of a is",a)
> 
>     for i in numbers:
>         print("Value of i is",i)
> 
>     for i, j in dict.items():
>         print("The value of i and j are:",i,j)
> 
> fun_varargs(1,2,3,4,5,6,7,8,9,10,Jack=111,John=222,Jimmy=333)
> 
> How do I make the tuple 'number'  contain the first element to be 1 and
> not 2?

One option is to change the function signature to

def fun_varargs(*numbers, a=5, **dict):
    ...

which turns `a` into a keyword-only argument.

>>> fun_varargs(1, 2, 3, foo="bar"):
Value of a is 5
Value of i is 1
Value of i is 2
Value of i is 3
The value of i and j are: foo bar

To override the default you have to specify a value like this:

>>> fun_varargs(1, 2, 3, foo="bar", a=42)
Value of a is 42
Value of i is 1
Value of i is 2
Value of i is 3
The value of i and j are: foo bar