Issues with script for adding numbers

Question

I am having issues submitting my script right now. I wonder if any of you have any ideas on why it is not taking it as it is working currently? It wants to add 6, 9, -4 and 7 together and my script is doing that.. Maybe there are ways to improve it? Any ideas would help out a lot.

#!/bin/bash

sum=0

for number in $1 $2 $3 $4
do
        sum=$(($1 + $2 + $3 + $4))
        echo $number

done

echo $sum

Answer 1

If you only want to add 4 numbers, this should be enough:

sum() {
  printf '%d\n' "$(( $1 + $2 + $3 + $4 ))"
}

Or if you want to add an arbitrary amount of numbers:

sum() {
  local number sum

  for number; do
    (( sum += number ))
  done

  printf '%d\n' "$sum"
}

Answer 2

Make the shell work for you. And use external utilities when called for.

#!/bin/bash
IFS=+
echo "$*" | bc

Answer 3

There is no need to loop if you know you're always going to get four integers on the command line:

#!/bin/sh

sum=$(( $1 + $2 + $3 + $4 ))
printf 'sum is %d\n' "$sum"

Alternatively, just

#!/bin/sh

printf 'sum is %d\n' "$(( $1 + $2 + $3 + $4 ))"

To support an arbitrary number of argument, you will need to loop:

#!/bin/sh

while [ "$#" -gt 0 ]; do
    sum=$(( sum + $1 ))
    shift
done

printf 'sum is %d\n' "$sum"

This script will iterate over its command line arguments, adding the first one to the sum variable and shifting it off the list of command line arguments, until no arguments are left. The $# variable expansion will expand to the number of command line arguments, while shift will remove $1 from the list, shifting $2 into its place (and $3 into $2 etc.).

Alternatively:

#!/bin/sh

for num do
    sum=$(( sum + num ))
done

printf 'sum is %d\n' "$sum"

Instead of constantly shifting the list of command line arguments, this leaves the list untouch and instead iterates over it, adding each one to sum in turn.

The for num do loop head may also be written for num in "$@"; do.