Call function in shell script

Question

I have the following code, that I want to be inside an function getsum(). I tried with the following code working without the function. When I run ./sum 5 6 I get 11.

#!/bin/bash
sum=0
for i in $@; do sum=$((sum+i)); done
echo $sum
exit 0

But how can I put it in a function doing the same job?

I tried the following code but it doesn't work.

#!/bin/bash
sums() {
    sum=0
    for i in $@; do sum=$((sum+i)); done
    echo $sum
    exit 0
}

sums

Apart from exit 0 your function looks fine. That statement will terminate your shell! — Elmar Peise
– Elmar Peise, Commented Oct 10, 2015 at 20:30

5gon12eder · Accepted Answer · 2015-10-10 20:39:01Z

3

You just need to pass the arguments ($@) to the function sum() that you pass to your script:

#!/bin/bash

sums() {
    sum=0
    for i in $@; do sum=$((sum+i)); done
    echo $sum
    exit 0
}

sums "$@" # Note this line

edited Oct 10, 2015 at 20:39

5gon12eder

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answered Oct 10, 2015 at 20:31

P.P

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