How to Increment and Decrement Variables in Bash

Introduction

Before we begin talking about how to increment and decrement variables in the Bash, let’s briefly understand - What does increment and decrement of variables mean?

Most common arithmetic operations when writing Bash scripts are incrementing and decrementing variables. These are most often useful for loops as a counter and can occur somewhere else in the script as well.

So, both incrementing and decrementing means adding or subtracting a value usually 1, respectively, from a numeric variable. The arithmetic expansion is performed using double parentheses ((...)) and $((...)) or, with the built-in let command. There are multiple ways to increment/decrement any variables.

In this tutorial, you will increment and decrement variable values in the Bash. We will also address some of the FAQs related to increment and decrement of variables.

Step 1 - Using + and - Operators

1) Given below is a most simple way to increment/decrement a variable. Do it by using the + and - operators.

i=$((i+1))
((i=i+1))
let "i=i+1"

i=$((i-1))
((i=i-1))
let "i=i-1"

This method will allow you to increment/decrement the variable by any value you want.

Below is an example of incrementing a variable with an until loop:

i=0

until [ $i -gt 3 ]
do
  echo i: $i
  ((i=i+1))
done

Output

i: 0
i: 1
i: 2
i: 3

Step 2 - Both += and -= Operators

1) In addition to basic operators explained above, bash also gives assignment operators += and -=. These operators are useful to increment/decrement the value of the left operand with value specified after the operator:

((i+=1))
let "i+=1" 

((i-=1))
let "i-=1" 

2) You can decrement the value of i variable by 5 with while loop as shown below:

i=20

while [ $i -ge 5 ]
do
  echo Number: $i
  let "i-=5" 
done

Output

Number: 20
Number: 15
Number: 10
Number: 5

Step 3 - Using ++ and -- Operators

1) Both ++ and -- operators increment and decrement, respectively, it's operand by 1 and returns the value as follows:

((i++))
((++i))
let "i++"
let "++i"

((i--))
((--i))
let "i--"
let "--i"

2) The operators are useful before or after the operand. They are even known as:

• The prefix increment- ++i
• Prefix decrement-  --i
• Postfix increment-  i++
• and, Postfix decrement-  i--

The prefix operators firstly increment/decrement operators by 1 and after that return the new value of operators. On the other hand, postfix operators return the operator's value before it has been incremented or decremented.

There is no difference in using prefix/postfix operator if you only want to increment/decrement the variables. It will make a difference only when result of the operators is useful in some other operation or assignable to another variable.

The below examples demonstrate how ++ operator works. Especially when it is useful before and after its operant:

x=5
y=$((x++))
echo x: $x
echo y: $y

Output

x: 6
y: 5

x=5
y=$((++x))
echo x: $x
echo y: $y

Output

x: 6
y: 6

3) Below is an example on the way to use postfix incrementor in a bash script:

#!/bin/bash
i=0
while true; do
  if [[ "$i" -gt 3 ]]; then
       exit 1
  fi
  echo i: $i
  ((i++))
done

There is a disadvantage in using these operators. These variables can only be incremented or decremented by 1.

FAQs to Increment and Decrement Variables in Bash

Conclusion

We hope this detailed guide helped you to increment and decrement variables in the Bash.

If you have any queries or doubts, please leave them in the comment below. We'll be happy to address them.