Integer variables in Bash Scripting - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

Output of integer arithmetic in Bash

What is the output of this Bash script?

Bash Scripting

a=5
b=3
c=$((a + b * 2))
echo $c

A 11

B 16

C 13

D 8

💡 Hint

Remember that multiplication has higher precedence than addition.

In Bash arithmetic, multiplication happens before addition. So b * 2 = 6, then a + 6 = 11.

📝 Syntax

intermediate

Correct integer assignment in Bash

Which option correctly assigns the integer value 10 to variable x in Bash?

A x=10

B x := 10

C x == 10

D let x = 10

💡 Hint

In Bash, assignment does not use spaces around = and no extra symbols.

Bash assigns variables with no spaces around =. 'x=10' is correct. 'let' is for arithmetic evaluation, not assignment. ':=' and '==' are invalid here.

🔧 Debug

advanced

Why does this Bash script fail to add integers?

Given this script: x=5 y=3 z=$x + $y echo $z What is the output and why?

Bash Scripting

x=5
y=3
z="$x + $y"
echo $z

A 53

B 8

C Syntax error

D 5 + 3

💡 Hint

Check how arithmetic is done in Bash and how variables are expanded.

The expression '$x + $y' is treated as a string, not arithmetic. So z becomes '5 + 3'. To do arithmetic, use $((x + y)).

🚀 Application

advanced

Calculate factorial using integer variables in Bash

Which script correctly calculates factorial of 4 and prints the result?

A

n=4
fact=1
while [ $n -gt 0 ]; do
  fact=$((fact * n))
  n=$((n - 1))
done
echo $fact

B

n=4
fact=1
for ((i=1; i<=n; i++)); do
  fact=$((fact * i))
done
echo $fact

C

n=4
fact=1
for i in $(seq 1 $n); do
  fact=$((fact + i))
done
echo $fact

D

n=4
fact=1
for i in {1..n}; do
  fact=$((fact * i))
done
echo $fact

💡 Hint

Check the loop syntax and arithmetic operation used for factorial.

Option B uses correct C-style for loop and multiplies fact by i each iteration. B fails because {1..n} does not expand n variable. A counts down but is correct factorial logic, but n is modified so loop ends correctly. D adds instead of multiplies.

🧠 Conceptual

expert

Integer variable behavior in Bash with arithmetic expansion

Consider this Bash snippet: x=10 y=3 z=$((x / y)) echo $z What is the output and why?

A 3.3333 because Bash supports floating point division

B 4 because Bash rounds up the division result

C 3 because Bash integer division truncates the decimal part

D Error because division is not supported in Bash arithmetic

💡 Hint

Bash arithmetic expansion works with integers only.

Bash integer arithmetic truncates results, so 10 / 3 equals 3, not 3.3333 or 4. Bash does not support floating point in $(( )).

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