Integer variables in Bash Scripting - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an integer variable in bash scripting?

An integer variable in bash is a variable that stores whole numbers without decimals. It is used to perform arithmetic operations easily.

beginner

How do you declare an integer variable in bash?

You declare an integer variable by using the keyword declare -i followed by the variable name. For example: declare -i count=5.

intermediate

What happens if you assign a non-integer value to an integer variable in bash?

Bash will try to convert the value to an integer. If it cannot, the variable will be set to 0.

beginner

How can you perform arithmetic operations with integer variables in bash?

You can use arithmetic expansion like $(( expression )). For example: sum=$((a + b)) adds two integer variables.

beginner

Why use integer variables instead of regular variables for numbers in bash?

Integer variables ensure that arithmetic operations work correctly and prevent errors from treating numbers as strings.

How do you declare an integer variable named 'num' with value 10 in bash?

A int num=10

B declare -i num=10

C num=10

D var num=10

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

What will be the value of an integer variable if you assign 'abc' to it?

A 0

B abc

C null

D error

Bash converts non-integer assignments to 0 for integer variables.

Which syntax correctly adds two integer variables 'a' and 'b' in bash?

A sum=$((a + b))

B sum = $(a + b)

C sum = $a + $b

D sum = a + b

Arithmetic expansion $(( )) is used for calculations in bash.

What keyword is used to declare an integer variable in bash?

A int

B integer

C declare -i

D var

declare -i is the bash keyword for integer variables.

Why is it better to use integer variables for numbers in bash scripts?

A To make scripts run faster

B To store text values

C To save memory

D To perform arithmetic correctly

Integer variables ensure arithmetic operations work properly.

Explain how to declare and use integer variables in bash scripting.

Think about how bash treats numbers and how to do math with variables.

Describe why integer variables are important in bash scripts and how they differ from regular variables.

Consider what happens when you do math with strings vs integers.

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