Shell Script: How do I add the digits of a number?

Question

I am making a shell script that takes a single number (length is unimportant) from the command line and adds the digits of it together. I thought I had it, but it won't work and either displays "0+3+4+5" if the command input is 345 or it displays the variables when I use expr to add them.

#!/bin/bash
sum=0
i="$(expr length $1)"
s=$1

for i in $(seq 0 $((${#s} - 1))); do
    value=${s:$i:1}
    typeset -i value
    sum=$sum+$value
done
echo $sum

Also doesn't work when I replace it with sum='expr $sum + $value'

any ideas?

dragonroot · Accepted Answer · 2011-11-03 03:22:28Z

4

What you are looking for is sum=$(($sum+$value)).

answered Nov 3, 2011 at 3:22

dragonroot

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Keith Thompson · Accepted Answer · 2011-11-03 04:01:04Z

0

#!/bin/bash

expr $(echo $1| sed 's/./& + /g;s/..$//')

For example, if the argument is 12345, this translates it to the string 1 + 2 + 3 + 4 + 5 and uses expr to evaluate it.

answered Nov 3, 2011 at 4:01

Keith Thompson

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