Subtract two variables in Bash

Question

Try this Bash syntax instead of trying to use an external program expr:

count=$((FIRSTV-SECONDV))

BTW, the correct syntax of using expr is:

count=$(expr $FIRSTV - $SECONDV)

But keep in mind using expr is going to be slower than the internal Bash syntax I provided above.

You just need a little extra whitespace around the minus sign, and backticks:

COUNT=`expr $FIRSTV - $SECONDV`

Be aware of the exit status:

The exit status is 0 if EXPRESSION is neither null nor 0, 1 if EXPRESSION is null or 0.

Keep this in mind when using the expression in a bash script in combination with set -e which will exit immediately if a command exits with a non-zero status.

You can use:

((count = FIRSTV - SECONDV))

to avoid invoking a separate process, as per the following transcript:

pax:~$ FIRSTV=7
pax:~$ SECONDV=2
pax:~$ ((count = FIRSTV - SECONDV))
pax:~$ echo $count
5

This is how I always do maths in Bash:

count=$(echo "$FIRSTV - $SECONDV"|bc)
echo $count

White space is important, expr expects its operands and operators as separate arguments. You also have to capture the output. Like this:

COUNT=$(expr $FIRSTV - $SECONDV)

but it's more common to use the builtin arithmetic expansion:

COUNT=$((FIRSTV - SECONDV))

For simple integer arithmetic, you can also use the builtin let command.

 ONE=1
 TWO=2
 let "THREE = $ONE + $TWO"
 echo $THREE
    3

For more info on let, look here.

Alternatively to the suggested 3 methods you can try let which carries out arithmetic operations on variables as follows:

let COUNT=$FIRSTV-$SECONDV

or

let COUNT=FIRSTV-SECONDV