Subtract two variables in Bash

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

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.

This is how I always do maths in Bash:

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

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.