Integer variables in Bash Scripting - Mini Project: Build & Apply

Integer variables

📖 Scenario: You are writing a simple bash script to store and use integer numbers. This is like keeping track of counts or scores in a game.

🎯 Goal: Learn how to create integer variables in bash, assign values, and display them.

📋 What You'll Learn

Create integer variables with exact names and values

Use bash syntax for integer assignment

Print the integer variable values correctly

💡 Why This Matters

🌍 Real World

Integer variables are used in scripts to count items, track progress, or calculate values automatically.

💼 Career

Knowing how to handle numbers in bash scripts is important for automation tasks in system administration and DevOps.

1

Create integer variables

Create two integer variables called count and max. Set count to 5 and max to 10 using bash integer assignment syntax.

Bash Scripting

# Create integer variables count and max with values 5 and 10
# Your code here

Hint

Use variable=value without spaces for integer assignment in bash.

2

Add a limit variable

Add a new integer variable called limit and set it to 15.

Bash Scripting

count=5
max=10
# Add integer variable limit with value 15
# Your code here

Hint

Remember no spaces around the equal sign.

3

Calculate total

Create a new integer variable called total that adds count and max using bash arithmetic expansion.

Bash Scripting

count=5
max=10
limit=15
# Calculate total as count + max
# Your code here

Hint

Use $((expression)) to do arithmetic in bash.

4

Print the total

Print the value of the total variable using echo.

Bash Scripting

count=5
max=10
limit=15
total=$((count + max))
# Print total
# Your code here

Hint

Use echo $total to display the value.

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