Function which returns Sum of all arguments

Question

I want bo create a function which returns sum of all passed arguments.

I found similar problem on forum, but It doesn't work with last case.

def sum_all(*args):
    sum = 0

    for num in args:
        if not str(num).isdigit():
            return False
        else:
            sum += int(num)
    return sum

This code work with first and second case, but return false on third case.

sum_all(2,-3)
# -1
sum_all([3,4,5])
# 12
sum_all(1,2,[3,4,5])
# false

How can i make it work? To make it return 15 in last case? Thanks!

You need to flatten `*args`: https://stackoverflow.com/questions/952914/how-to-make-a-flat-list-out-of-list-of-lists — Maurice Meyer, Dec 07 '20 at 16:03

score 2 · Accepted Answer · answered Dec 07 '20 at 16:11

2

In case of a list, just call the function recursively, example:

def sum_all(*args):
  sum = 0
  for num in args:
    if type(num) is list:
      sum += sum_all(*num)
    else:
      sum += int(num)
  return sum

answered Dec 07 '20 at 16:11

Simona

349
2
8

Don't you still need a check for an input that isn't numeric? – Barmar Dec 07 '20 at 16:24
this is simplified, I didn't do entire error handling. I just demonstrated the needed recursion – Simona Dec 07 '20 at 16:26

score 1 · Answer 2 · answered Dec 07 '20 at 16:17

1

Can use this

def sum_all(*args):
    s = 0
    for a in args:
        if type(a)==list:
            s+=sum(a)
        else:
            s+=int(a)
    return s
print(sum_all(2,-3))
print(sum_all([3,4,5]))
print(sum_all(1,2,[3,4,5]))

answered Dec 07 '20 at 16:17

theCoder

34
5

Function which returns Sum of all arguments

2 Answers2