Bash script to add absolute values of numbers seperated by spaces

Question

I need a bash script to find the sum of the absolute value of integers separated by spaces. For instance, if the input is:

1 2 -3

the script should print 6 to standard output I have:

while read x ; do echo $(( ${x// /+} )) ; done

which gives me

Without over complicated things, how would I include an absolute value of each x in that statement so the output would be:

You could remove all `-` characters before converting the spaces to `+`. — Barmar, Nov 08 '21 at 23:32

score 4 · Answer 1 · answered Nov 08 '21 at 23:39

4

With Barmar's idea:

echo "1 2 -3" | tr -d - | tr ' ' '+' | bc -l

Output:

answered Nov 08 '21 at 23:39

Cyrus

can you show me your example using a variable instead of 1 2 -3 hard coded? – Steve Zee Nov 08 '21 at 23:44
Replace `1 2 -3` with `$x`. – Cyrus Nov 08 '21 at 23:54

score 3 · Accepted Answer · answered Nov 09 '21 at 10:36

3

You have almost done it, but the -s must have been removed from the line read:

while read x; do x=${x//-}; echo $(( ${x// /+} )); done

answered Nov 09 '21 at 10:36

M. Nejat Aydin

score 2 · Answer 3 · answered Nov 09 '21 at 00:42

2

POSIX friendly implementation without running a loop and without spawning a sub-shell:

#!/usr/bin/env sh

abssum() {
  IFS='-'
  set -- $*
  IFS=' '
  set -- $*
  IFS=+
  printf %d\\n $(($*))
}

abssum 1 2 -3

Result:

answered Nov 09 '21 at 00:42

Léa Gris

3 Answers3