Bash script to sum absolute values?

Question

Is it possible to sum absolute values for integers entered into stdin and print the result to stdout?

So far I have is this:

read X
read Y
echo "$(($X+$Y))"

Can return:

11
11
22

But looking to get a result for absolutes. E.g:

-11
11
22

See this similar question https://stackoverflow.com/questions/29223313/absolute-value-of-a-number — MattCochrane, May 15 '20 at 07:56
Does this answer your question? [Absolute value of a number](https://stackoverflow.com/questions/29223313/absolute-value-of-a-number) — MattCochrane, May 15 '20 at 07:57

score 2 · Accepted Answer · answered May 15 '20 at 08:05

Consider modified version, working on modern bash.

With temporary variable - note expression on let must be quoted, otherwise the special characters '<' will break the expression logic.

read x
read y
let s="(x<0?-x:x)+(y<0?-y:y)"
echo "SUM(abs($x)+abs($y))=$s"

Or without intermediate variable, using the '$((expr))' grammar. No need to quote.

read x
read y
echo "SUM(abs($x)+abs($y))=" "$(((x<0?-x:x)+(y<0?-y:y)))"

score 1 · Answer 2 · answered May 15 '20 at 12:19

1

Just strip the leading - from the variables:

read x
read y
echo "$((${x#-} + ${y#-}))"

This works since numbers in Bash are represented as strings.

Note that you should validate the user input (so the sign removal could be done there).

answered May 15 '20 at 12:19

gniourf_gniourf

2 Answers2