CODEWITHSUNDEEP
bashfunctionshell
Alex
Alex

Reputation: 101

Call function in shell script

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

Upvotes: 0

Views: 63

Answers (1)

P.P
P.P

Reputation: 121347

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

Upvotes: 3

Related Questions