Python Variable Argument to functions

Question

I am learning Python and came across variadic arguments. I don't understand the output that the following code produces:

_list = [11,2,3]
def print_list(*args):
    for value in args:
        a = value * 10
        print(a)
print_list(_list)

When I run the program, I get:

[11, 2, 3, 11, 2, 3, 11, 2, 3, 11, 2, 3, 11, 2, 3, 11, 2, 3, 11, 2, 3, 11, 2, 3, 11, 2, 3, 11, 2, 3]

From what I understand, value holds one single element from the _list array, multiplying it by 10 would produce the list [110, 20, 30]. Why does the output differ?

try doing print(args) and I think you'll see why. args in your case is ([11,2,3],) -- a length 1 tuple of lists. — FHTMitchell
– FHTMitchell, Commented Aug 22, 2018 at 9:59

khelwood · Accepted Answer · 2018-08-22 10:00:30Z

6

Because the parameter to your function is *args (with a *), your function actually receives a tuple of the passed in arguments, so args becomes ([11,2,3],) (a tuple containing the list you passed in).

Your function iterates through the value in that tuple, giving value=[11,2,3]. When you multiply a list by 10, you get a list 10 times longer.

answered Aug 22, 2018 at 10:00

khelwood

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1 Comment

Tesla Over a year ago

Got it, Thank you... Coming from the C++ background, I mistook *args with pointer-like functionality. Will learn tuples in python now.

Jirka · Accepted Answer · 2018-08-27 17:28:20Z

0

Probably what are you looking for is expanding the list print_list(*_list) which pass each element of the input list as a function parameter and so the output is 110 20 30.

If you would have the list as numpy.array(_list) * 10 then the multiplications will also work.

edited Aug 27, 2018 at 17:28

answered Aug 22, 2018 at 10:23

Jirka

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