Linux shell negative values print problem

Question

operation () {
   operator=${exp:1:1} # 2nd character / 1st is ${words:0:1} 

   # Read into an array as tokens separated by IFS
   IFS="$operator"
   read -ra array <<< "$exp" # -ra = raw input array
   a=array[0]
   b=array[1]
   # https://www.computerhope.com/unix/bash/read.htm

   if [ "$operator" == "+" ]; then
    operation=$((a+b))
   fi
   if [ "$operator" == "-" ]; then
    operation=$((a+b)) # => ' 1'
    operation=`expr a-b` # => ' 1'
   fi
   if [ "$operator" == "*" ]; then
    operation=$((a*b))
   fi
   if [ "$operator" == "/" ]; then
    operation=$((a/b))
   fi
   # https://www.tutorialsandyou.com/bash-shell-scripting/bash-arithmetic-operations-11.html
   echo $((operation))
}
exp='2-3'
operation $exp 

Everything(almost) works fine, just when making the subtraction operation=$((a+b)) or operation=`expr ab` I cannot print in anyway minus('-') instad of space(' '), the (-) sign it is been replaced by space. In this example I get ' 1' instead of '-1' .

Why is this happening?

Answer 1

use space around -:

exp=' 2 - 3 '

or

operation = expr a - b

or

COUNT="expr $FIRSTV - $SECONDV"

also see this [ link ] 1

Answer 2

I guess you meant to use - if operator is a minus sign:

if [ "$operator" == "-" ]; then
    operation=$((a - b)) # => ' 1'
fi

There are 2 problems with this script. First, arrays elements are not referenced like array[0] is in C but ${name[subscript]} as it says in man bash:

Any element of an array may be referenced using ${name[subscript]}.

So it should be:

a=${array[0]}
b=${array[1]}

Second, $((var)) should only be used for arithmetic. If you want to print contents of variable just do:

echo "$operation"

All in all, your script should be:

#!/usr/bin/env bash

operation () {
    operator=${exp:1:1} # 2nd character / 1st is ${words:0:1}

    # Read into an array as tokens separated by IFS
    IFS="$operator"
    read -ra array <<< "$exp" # -ra = raw input array
    a=${array[0]}
    b=${array[1]}
    # https://www.computerhope.com/unix/bash/read.htm

    if [ "$operator" == "+" ]; then
        operation=$((a+b))
    fi
    if [ "$operator" == "-" ]; then
        operation=$((a - b))
    fi
    if [ "$operator" == "*" ]; then
        operation=$((a*b))
    fi
    if [ "$operator" == "/" ]; then
        operation=$((a/b))
    fi
    # https://www.tutorialsandyou.com/bash-shell-scripting/bash-arithmetic-operations-11.html
    echo "$operation"
}
exp='2-3'
operation $exp

Usage:

$ ./script.sh
-1

Answer 3

We're missing some context here:

• This a=array[0] is only correct if declare -ia .
• What is exp ?

But the problem is a simple matter of lack of quoting:

$ IFS="-"
$ operation="-6"
$ echo $((operation))
 6
$ echo "$((operation))"
-6

This is entirely due to the IFS value. bash gets echo -6 and translates that into echo "" 6 with the empty string to the left of the - delimiter.

Within quotes, word splitting is not performed.

Behaviour documented in 3.5.7 Word Splitting in the manual.

---

You can set IFS only for the duration of the read command so it does not affect the rest of the script:

$ exp="4-5"
$ IFS=- read -ra array <<<"$exp"
$ declare -p array
declare -a array=([0]="4" [1]="5")
$ printf "%q\n" "$IFS"
$' \t\n'

Answer 4

1. I remove this line operation='expr ab' # => ' 1'
2. In this line operation=$((a+b)) # => ' 1' you must use - instead +:
3. For output use "" around the variables :

echo "$((operation))"

This code work properly:

operation () {
   operator=${exp:1:1} # 2nd character / 1st is ${words:0:1} 

   # Read into an array as tokens separated by IFS
   IFS="$operator"
   read -ra array <<< "$exp" # -ra = raw input array
   a=array[0]
   b=array[1]
   
   # https://www.computerhope.com/unix/bash/read.htm

   if [ "$operator" == "+" ]; then
    operation=$((a+b))
   fi
   if [ "$operator" == "-" ]; then
    operation="$((a-b))"
   fi
   if [ "$operator" == "*" ]; then
    operation=$((a*b))
   fi
   if [ "$operator" == "/" ]; then
    operation=$((a/b))
   fi
   # https://www.tutorialsandyou.com/bash-shell-scripting/bash-arithmetic-operations-11.html
   echo "$((operation))"
}
exp='2-3'
operation $exp