Subtract two variables in Bash - bash

Question

I have the script below to subtract the counts of files between two directories but the COUNT= expression does not work. What is the correct syntax?
#!/usr/bin/env bash
FIRSTV=`ls -1 | wc -l`
cd ..
SECONDV=`ls -1 | wc -l`
COUNT=expr $FIRSTV-$SECONDV ## -> gives 'command not found' error
echo $COUNT

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

This is how I always do maths in Bash:
count=$(echo "$FIRSTV - $SECONDV"|bc)
echo $count

Answer 5

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))

Answer 6

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.

Answer 7

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

Answer 8

Diff Real Positive Numbers
diff_real () {
echo "df=($1 - $2); if (df < 0) { df=df* -1}; print df" | bc -l;
}
Usage
var_a=10
var_b=4
output=$(diff_real $var_a $var_b)
# 6
#########
var_a=4
var_b=10
output=$(diff_real $var_a $var_b)
# 6

Answer 9

Use BASH:
#!/bin/bash
# home/victoria/test.sh
START=$(date +"%s") ## seconds since Epoch
for i in $(seq 1 10)
do
sleep 1.5
END=$(date +"%s") ## integer
TIME=$((END - START)) ## integer
AVG_TIME=$(python -c "print(float($TIME/$i))") ## int to float
printf 'i: %i | elapsed time: %0.1f sec | avg. time: %0.3f\n' $i $TIME $AVG_TIME
((i++)) ## increment $i
done
Output
$ ./test.sh
i: 1 | elapsed time: 1.0 sec | avg. time: 1.000
i: 2 | elapsed time: 3.0 sec | avg. time: 1.500
i: 3 | elapsed time: 5.0 sec | avg. time: 1.667
i: 4 | elapsed time: 6.0 sec | avg. time: 1.500
i: 5 | elapsed time: 8.0 sec | avg. time: 1.600
i: 6 | elapsed time: 9.0 sec | avg. time: 1.500
i: 7 | elapsed time: 11.0 sec | avg. time: 1.571
i: 8 | elapsed time: 12.0 sec | avg. time: 1.500
i: 9 | elapsed time: 14.0 sec | avg. time: 1.556
i: 10 | elapsed time: 15.0 sec | avg. time: 1.500
$
Get current time in seconds since the Epoch on Linux, Bash