Nested conditional statements - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a number is positive.

C

if (num [1] 0) {
    printf("Positive number\n");
}

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' will check for negative numbers.

Using '==' checks for equality, not positivity.

The condition 'num > 0' checks if the number is positive.

2fill in blank

medium

Complete the code to check if a number is negative.

C

if (num [1] 0) {
    printf("Negative number\n");
}

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

A>

B!=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' will check for positive numbers.

Using '==' checks for equality, not negativity.

The condition 'num < 0' checks if the number is negative.

3fill in blank

hard

Fix the error in the nested if statement to check if a number is zero.

C

if (num > 0) {
    printf("Positive\n");
} else if (num [1] 0) {
    printf("Zero\n");
} else {
    printf("Negative\n");
}

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

A!=

B<

C==

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '<' instead of '==' will not correctly check for zero.

Using '!=' checks for not equal, which is incorrect here.

To check if num is zero, use 'num == 0' in the else if condition.

4fill in blank

hard

Fill both blanks to complete the nested if-else that prints if a number is positive even or odd.

C

if (num [1] 0) {
    if (num [2] 2 == 0) {
        printf("Positive even\n");
    } else {
        printf("Positive odd\n");
    }
}

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

A>

B<

C%

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' for positivity check.

Using '==' instead of '%' for even check.

First check if num > 0, then use 'num % 2 == 0' to check evenness.

5fill in blank

hard

Fill all three blanks to complete the nested if-else that categorizes a number as positive even, positive odd, or non-positive.

C

if (num [1] 0) {
    if (num [2] 2 [3] 0) {
        printf("Positive even\n");
    } else {
        printf("Positive odd\n");
    }
} else {
    printf("Non-positive number\n");
}

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

A>

B%

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '>' for positivity check.

Using '!=' instead of '==' for even check.

Check if num > 0, then use 'num % 2 == 0' to check evenness, else non-positive.

Practice

1.

What does a nested conditional statement in C allow you to do?

Choose the best description.

easy

A. Define a function inside another function

B. Check a condition inside another condition

C. Repeat a block of code multiple times

D. Create a variable that changes type

Solution

1. Step 1: Understand the meaning of nested conditionals

   Nested conditionals mean putting one if or else if statement inside another if or else if block.

2. Step 2: Compare with other options

   Repeating code is done by loops, not conditionals. Functions inside functions or variable type changes are unrelated.

3. Final Answer:

   Check a condition inside another condition -> Option B

4. Quick Check:

   Nested conditionals = condition inside condition [OK]

Hint: Nested means inside another condition [OK]

Common Mistakes:

• Confusing nested conditionals with loops
• Thinking nested means changing variable types
• Mixing nested conditionals with function definitions

2.

Which of the following is the correct syntax for a nested if statement in C?

if (x > 0) {
    if (y < 10) {
        // code
    }
}

easy

A. if x > 0 { if (y < 10) { /* code */ } }

B. if (x > 0) if (y < 10) { /* code */ }

C. if (x > 0) { if y < 10 { /* code */ } }

D. if (x > 0) { if (y < 10) { /* code */ } }

Solution

1. Step 1: Check correct use of braces and parentheses

   In C, conditions must be inside parentheses and blocks inside braces. if (x > 0) { if (y < 10) { /* code */ } } uses both correctly.

2. Step 2: Identify syntax errors in other options

   if (x > 0) if (y < 10) { /* code */ } misses braces for outer if block. if (x > 0) { if y < 10 { /* code */ } } misses parentheses around y < 10. if x > 0 { if (y < 10) { /* code */ } } misses parentheses around x > 0.

3. Final Answer:

   if (x > 0) { if (y < 10) { /* code */ } } -> Option D

4. Quick Check:

   Correct syntax = if (x > 0) { if (y < 10) { /* code */ } } [OK]

Hint: Use parentheses for conditions and braces for blocks [OK]

Common Mistakes:

• Omitting parentheses around conditions
• Forgetting braces for code blocks
• Mixing syntax from other languages

3.

What is the output of this C code?

int x = 5, y = 8;
if (x > 0) {
    if (y < 10) {
        printf("A\n");
    } else {
        printf("B\n");
    }
} else {
    printf("C\n");
}

medium

A. A

B. B

C. C

D. No output

Solution

1. Step 1: Evaluate outer if condition

   x = 5, which is greater than 0, so the outer if block runs.

2. Step 2: Evaluate inner if condition

   y = 8, which is less than 10, so the inner if block runs and prints "A".

3. Final Answer:

   A -> Option A

4. Quick Check:

   x > 0 and y < 10 prints A [OK]

Hint: Check outer condition first, then inner [OK]

Common Mistakes:

• Ignoring inner condition
• Confusing else blocks
• Assuming no output if outer condition true

4.

Find the error in this nested conditional code snippet:

if (a > 10)
    if (b < 5)
        printf("X\n");
    else
        printf("Y\n");

medium

A. Missing braces causing else to pair with wrong if

B. Incorrect use of semicolons after if statements

C. Using printf without including stdio.h

D. Variables a and b are not declared

Solution

1. Step 1: Analyze the if-else pairing

   Without braces, the else pairs with the nearest if, which is the inner if. This may cause confusion or unexpected behavior.

2. Step 2: Identify the fix

   Adding braces around the outer if block ensures the else pairs correctly and code is clear.

3. Final Answer:

   Missing braces causing else to pair with wrong if -> Option A

4. Quick Check:

   Braces fix else pairing [OK]

Hint: Always use braces to avoid else ambiguity [OK]

Common Mistakes:

• Assuming else pairs with outer if
• Ignoring braces importance
• Thinking semicolons cause this error

5.

Given the code below, what will be printed if score = 75?

if (score >= 90) {
    printf("Grade A\n");
} else {
    if (score >= 70) {
        if (score >= 80) {
            printf("Grade B+\n");
        } else {
            printf("Grade B\n");
        }
    } else {
        printf("Grade C\n");
    }
}

hard

A. Grade B+

B. Grade C

C. Grade B

D. Grade A

Solution

1. Step 1: Check first if condition

   score = 75, which is not >= 90, so skip first if block.

2. Step 2: Check second if condition

   score >= 70 is true, so enter second if block.

3. Step 3: Check nested if inside second block

   score >= 80 is false (75 < 80), so else block runs printing "Grade B".

4. Final Answer:

   Grade B -> Option C

5. Quick Check:

   75 >= 70 but < 80 prints Grade B [OK]

Hint: Check conditions top-down carefully [OK]

Common Mistakes:

• Confusing >= 70 and >= 80 checks
• Skipping nested else block
• Assuming 75 is Grade B+