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+$Y))"
Can return:
11
11
22
But looking to get a result for absolutes. E.g:
-11
11
22
2 Answers
Consider modified version, working on modern bash.
With temporary variable - note expression on let must be quoted, otherwise the special characters '<' will break the expression logic.
read x
read y
let s="(x<0?-x:x)+(y<0?-y:y)"
echo "SUM(abs($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)))"
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).
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