Using default arguments in a function with variable arguments. Is this possible?

Question

I have the following piece of code. In this, I want to make use of the optional parameter given to 'a'; i.e '5', and not '1'. How do I make the tuple 'numbers' contain the first element to be 1 and not 2?

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,Tom=333)

Answer 1

The "idiomatic" Python way to do this is to have the optional argument default to None.

def fun_varargs(a=None, *numbers, **dict):
  if a is None:
    a = 5
  ...

Then you can explicitly choose not to pass it.

fun_varargs(None, 1, 2, 3, 4, 5, some_keyword_arg=":)")

I don't know of a way to actually choose not to pass an optional argument. If you're looking for a practical solution, this is the way to go. But if there's some deeper hackery that lets you bypass the argument altogether, I'd be interested in seeing it as well.