Integer variables in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an integer variable named num with value 10.

Bash Scripting

declare -i [1]=10

Drag options to blanks, or click blank then click option'

Anum

Bnumber

Cint

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name different from 'num'.

Forgetting to declare the variable as integer.

The declare -i command declares an integer variable. Here, num is the variable name.

2fill in blank

medium

Complete the code to add 5 to the integer variable count.

Bash Scripting

count=[1]+5

Drag options to blanks, or click blank then click option'

Acount

B$count

C"count"

D'count'

Attempts:

3 left

💡 Hint

Common Mistakes

Using the variable name without $.

Using quotes around the variable name.

To use the current value of count, prefix it with $ to get its value.

3fill in blank

hard

Fix the error in the code to correctly increment the integer variable score by 1.

Bash Scripting

(( [1]++ ))

Drag options to blanks, or click blank then click option'

A'score'

B$score

C"score"

Dscore

Attempts:

3 left

💡 Hint

Common Mistakes

Using $score inside arithmetic expression.

Using quotes around the variable name.

Inside arithmetic expressions, use the variable name without $ or quotes.

4fill in blank

hard

Fill in the blanks to declare an integer variable total and assign it the sum of a and b.

Bash Scripting

declare -i [1]=[2]+[3]

Drag options to blanks, or click blank then click option'

Atotal

Ba

Cb

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable name for declaration.

Using the same variable twice instead of both a and b.

Declare total as integer and assign it the sum of a and b. Both a and b are variables used in the expression.

5fill in blank

hard

Fill all three blanks to create a loop that counts from 1 to 5 and prints each number.

Bash Scripting

for [1] in [2]; do
  echo [3]
done

Drag options to blanks, or click blank then click option'

Ai

B1 2 3 4 5

C$i

Dseq 1 5

Attempts:

3 left

💡 Hint

Common Mistakes

Using the variable name without $ in echo.

Using seq 1 5 without command substitution.

Using quotes around the list of numbers.

Use i as the loop variable, iterate over the list 1 2 3 4 5, and print the value of i with $i.

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