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 $COUNT Try this Bash syntax instead of trying to use an external program expr:
count=$((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.
$(command) syntax for command substitution. Also since BASH support arithmetic operations in $(( ... )) it is better to not to use an external utility expr
$(( ... )) is use for evaluating arithmetic expressions.
$(( is not a bashism. It works in dash, and is in POSIX. Moreover, shellcheck will complain if you use expr: shellcheck.net/wiki/SC2003You just need a little extra whitespace around the minus sign, and backticks:
COUNT=`expr $FIRSTV - $SECONDV` Be aware of the exit status:
The exit status is 0 if EXPRESSION is neither null nor 0, 1 if EXPRESSION is null or 0.
Keep this in mind when using the expression in a bash script in combination with set -e which will exit immediately if a command exits with a non-zero status.
sh shell. For portability, you may want to use this answer.
expr only makes sense if you need compatibility back to the dark ages when Solaris sh was popular and HP-UX and AIX were still a thing. POSIX sh too has supported arithmetic expressions for many years now.You can use:
((count = FIRSTV - SECONDV)) to avoid invoking a separate process, as per the following transcript:
pax:~$ 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 $count 
|bc type command than miss it once or twice. Different strokes for different folks as they say.
White space is important, expr expects its operands and operators as separate arguments. You also have to capture the output. Like this:
COUNT=$(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 $THREE 3 For more info on let, look here.
let "sanity_check_duration=sanity_check_duration_end_time_delay_sec - sanity_check_duration_start_time_delay_sec" (removing dollar sign from variables)Alternatively to the suggested 3 methods you can try let which carries out arithmetic operations on variables as follows:
let COUNT=$FIRSTV-$SECONDV
or
let COUNT=FIRSTV-SECONDV
diff_real () { echo "df=($1 - $2); if (df < 0) { df=df* -1}; print df" | bc -l; } var_a=10 var_b=4 output=$(diff_real $var_a $var_b) # 6 ######### var_a=4 var_b=10 output=$(diff_real $var_a $var_b) # 6 Use BASH:
#!/bin/bash # home/victoria/test.sh START=$(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 $ 
