Arithmetic syntax in bash

Question

I am getting the below error on arithmetic values

#!/bin/bash
n=0
line_count=9
line_count=$(line_count)/3
echo $line_count
exit 0

expected result is 3

[]$ ./test.sh
./test.sh: line 4: line_count: command not found
/3
[]$ more test.sh

Answer 1

To complement @Kusalananda's answer, in addition to the standard sh syntax:

line_count=$((line_count / 3))

In bash you can also use these syntaxes inherited from ksh (also work in zsh):

• ((line_count = line_count / 3))

• ((line_count /= 3))

• let line_count/=3

• typeset -i line_count; line_count=line_count/3

bash (and zsh) also support:

• line_count=$[line_count/3]

For old pre-POSIX Bourne/Almquist sh:

line_count=`expr "$line_count" / 3`

Answer 2

Arithmetic expansion is done by bash and some other shells with $(( ... )), e.g.

line_count=$(( line_count/3 ))

With line_count=$(line_count)/3, you are assigning the output of the command line_count to the variable line_count, suffixed by /3.

This is why you get the error "line_count: command not found" and then the output /3.

Have a look at ShellCheck at https://www.shellcheck.net/ It is able to check you script for common issues. In this case, it won't help you with spotting that $(...) should have been $((...)) but that's because $(line_count) is perfectly legal shell code. It just does the wrong thing.