pass value to function with * and/or **

[buran]

if the caller does use * and/or **, does what the caller provides get passed through to the function as-is, or does it go through the argument handling mechanisms that give the function a tuple and dictionary

I think I show all possible variations in my example. You can do the same with your own type/class objects and see what the result is. Here it is again

from collections import OrderedDict

def foo(*boo, **hoo):
    print(type(boo), boo)
    print(type(hoo), hoo)
 
spam = [1, 2]
eggs = OrderedDict((('a', 3), ('b', 4)))
 
foo(*spam, **eggs)
foo(spam, eggs)
foo(5, 6, c=7, d=8)

Output:
<class 'tuple'> (1, 2)
<class 'dict'> {'a': 3, 'b': 4}
<class 'tuple'> ([1, 2], OrderedDict([('a', 3), ('b', 4)]))
<class 'dict'> {}
<class 'tuple'> (5, 6)
<class 'dict'> {'c': 7, 'd': 8}

as you can see that when unpacked elements of the list are stored in a tuple and elements of the OrderedDict are stored in a dict
when not unpacked you get list and OrderedDict as elements in a tuple.
bottom line - args/boo will be a tuple always, kwargs/hoo will be a dict always.