Use double bracket for arithmetic operation in bash

[mhdzumair]

Description:
Sir we can use double bracket for arithmetic operation in bash is more flexible way because it doesn't matter if there is a bracket or not. So I'm kindly suggesting to replace / include this double bracket guide.

Suggested Labels:
Improvement

Suggested Assignees:
Affected Product Version:
OS, DB, other environment details and versions:
Steps to reproduce:
Related Issues:

Current Code:

#!/bin/sh

A=70
B=90

val=`expr $A + $B` #Be Mindful about the spaces in between the "expr" , $A , + , $B  characters, 
#                  which  is a must. Otherwise script will throw a syntax error 
echo "A + B : $val"

val=`expr $A - $B`
echo "A - B : $val"

val=`expr $A \* $B`
echo "a * b : $val"

val=`expr $B / $A`
echo "B / A : $val"

val=`expr $B % $A`
echo "B % A : $val”

Suggesting Code

#!/usr/bin/bash

A=70
B=90

((val = $A + $B))
echo "A + B : $val"

((val=$A-$B))
echo "A - B : $val"

((val= $A * $B))
echo "a * b : $val"

((val = $B / $A))
echo "B / A : $val"

((val= $B % $A))
echo "B % A : $val"