[Tutor] Defining variable arguments in a function in python

[Steven D'Aprano]

On Sat, Dec 29, 2018 at 11:42:16AM +0530, 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):
[...]
> 
> 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?


You can't. Python allocates positional arguments like "a" first, and 
only then collects whatever is left over in *numbers. How else would you 
expect it to work? Suppose you called:

fun_varargs(1, 2, 3)

wanting a to get the value 1, and numbers to get the values (2, 3). And 
then immediately after that you call 

fun_varargs(1, 2, 3)

wanting a to get the default value 5 and numbers to get the values 
(1, 2, 3). How is the interpreter supposed to guess which one you 
wanted?

If you can think of a way to resolve the question of when to give "a" 
the default value, then we can help you program it yourself:


def func(*args, **kwargs):
    if condition:
        # When?
        a = args[0]
        numbers = args[1:]
    else:
        a = 5  # Default.
        numbers = args
    ...

But writing that test "condition" is the hard part.




-- 
Steve