Integer variables in Bash Scripting

Introduction

Integer variables store whole numbers so you can do math and count things in your script.

Counting how many files are in a folder.

Adding numbers like scores or totals.

Using loops that run a set number of times.

Tracking how many times something happens.

Doing simple math calculations in your script.

Syntax

Bash Scripting

declare -i variable_name=number
# or simply
variable_name=number

Use 'declare -i' to tell bash this variable is an integer.

You can assign a number directly without quotes.

Examples

This creates an integer variable named count with the value 10.

Bash Scripting

declare -i count=10

This assigns 5 to variable number. Bash treats it as a string but can do math if used properly.

Bash Scripting

number=5

Declares total as integer, sets it to 7, then adds 3 to it.

Bash Scripting

declare -i total
 total=7
 total=total+3

Sample Program

This script creates an integer variable score, adds points to it, and prints the final score.

Bash Scripting

#!/bin/bash

# Declare integer variable
declare -i score=0

# Add points
score=score+10
score=score+5

# Print the score
 echo "Your score is $score"

Important Notes

Without 'declare -i', bash treats variables as strings, so math needs special syntax like $(( )) to work.

Using 'declare -i' makes bash automatically calculate math expressions assigned to the variable.

Always use quotes around variables when printing to avoid word splitting or errors.

Summary

Integer variables hold whole numbers for math in bash scripts.

Use 'declare -i' to make a variable an integer.

You can add or change integer variables with simple math expressions.

Practice

1. What does the command declare -i num=5 do in a bash script?

easy

A. It creates an integer variable named num with value 5.

B. It creates a string variable named num with value '5'.

C. It declares a function named num that returns 5.

D. It prints the number 5 to the terminal.

Solution

1. Step 1: Understand declare -i usage

   The declare -i command in bash makes a variable hold integer values only.

2. Step 2: Analyze the assignment

   Here, num is assigned the integer value 5, so it becomes an integer variable.

3. Final Answer:

   It creates an integer variable named num with value 5. -> Option A

4. Quick Check:

   declare -i num=5 means integer variable num = 5 [OK]

Hint: declare -i means integer variable assignment [OK]

Common Mistakes:

• Thinking declare -i creates a string variable
• Confusing declare -i with printing output
• Assuming declare -i creates a function

2. Which of the following is the correct way to declare an integer variable count with initial value 10 in bash?

easy

A. count=10

B. declare count=10

C. declare -i count=10

D. int count=10

Solution

1. Step 1: Recall integer declaration syntax

   In bash, declare -i is used to declare integer variables.

2. Step 2: Check each option

   count=10 assigns a value but does not declare integer type. declare count=10 is invalid syntax. declare -i count=10 correctly uses declare -i. int count=10 is not valid bash syntax.

3. Final Answer:

   declare -i count=10 -> Option C

4. Quick Check:

   declare -i count=10 is correct integer declaration [OK]

Hint: Use declare -i to declare integer variables [OK]

Common Mistakes:

• Using declare without -i for integers
• Trying C-style int declaration in bash
• Assigning without declare -i loses integer behavior

3. What is the output of this bash script snippet?

declare -i x=7
declare -i y=3
x=$x+$y
echo $x

medium

A. 10

B. 7+3

C. 73

D. Error

Solution

1. Step 1: Understand variable declarations

   Both x and y are declared as integers with values 7 and 3.

2. Step 2: Analyze the assignment x=$x+$y

   Parameter expansion produces x=7+3. Since x is declare -i, bash performs arithmetic evaluation on the RHS: 7 + 3 = 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   $x+$y expands to 7+3, -i evaluates to 10 [OK]

Hint: declare -i evaluates valid arithmetic RHS like num+num [OK]

Common Mistakes:

• Thinking it concatenates to 73
• Expecting literal '7+3' output
• Believing explicit $(( )) always required

4. What is the output of this bash script snippet?

declare -i total=5
total=total+3
echo $total

medium

A. 5

B. 8

C. total+3

D. Error

Solution

1. Step 1: Check variable declaration

   total is declared as integer with value 5 [check].

2. Step 2: Analyze assignment total=total+3

   With declare -i, RHS total+3 is arithmetic context. Bare total expands to 5, evaluates 5+3=8.

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Bare var in -i RHS expands + computes math [OK]

Hint: Bare variable names expand in -i arithmetic assignments [OK]

Common Mistakes:

• Assuming literal 'total+3' without $
• Thinking no math without $(( ))
• Expecting unchanged 5

5. You want to create a bash script that counts from 1 to 5 using an integer variable i. Which snippet correctly increments i inside a loop?

hard

A. declare -i i=1
while [ $i -le 5 ]; do
echo $i
i=i +1
done

B. declare -i i=1
while [ $i -le 5 ]; do
echo $i
i++
done

C. declare i=1
while [ $i -le 5 ]; do
echo $i
i=$i+1
done

D. declare -i i=1
while [ $i -le 5 ]; do
echo $i
i=$((i+1))
done

Solution

1. Step 1: Understand integer variable and loop

   Declare i integer=1, loop while <=5, echo i, increment properly to print 1-5.

2. Step 2: Check increment methods

   i=i +1 parses as i=i (no change) then invalid +1 command, fails.
   i++ invalid command without (( )).
   No -i causes string concat 1+1->11 etc.
   i=$((i+1)) explicit arithmetic works portably.

3. Final Answer:

   i=$((i+1)) -> Option D

4. Quick Check:

   $((i+1)) for safe integer increment [OK]

Hint: Use i=$((i+1)) or ((i++)) to increment integers [OK]

Common Mistakes:

• Spaces in i=i +1 cause assignment split + command error
• i++ without (( )) is invalid command
• No -i leads to string concatenation