Nested conditional statements - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested if-else with integer input

What is the output of the following C code when num = 7?

C

#include <stdio.h>

int main() {
    int num = 7;
    if (num > 5) {
        if (num < 10) {
            printf("A\n");
        } else {
            printf("B\n");
        }
    } else {
        printf("C\n");
    }
    return 0;
}

A B

B C

C A

D No output

💡 Hint

Check the conditions step by step starting from the outer if.

The outer if checks if num > 5, which is true for 7. Then the inner if checks if num < 10, which is also true. So it prints 'A'.

❓ Predict Output

intermediate

Output of nested if-else with character input

What will be printed when the variable ch = 'b' is used in this code?

C

#include <stdio.h>

int main() {
    char ch = 'b';
    if (ch == 'a') {
        printf("Apple\n");
    } else {
        if (ch == 'b') {
            printf("Banana\n");
        } else {
            printf("Cherry\n");
        }
    }
    return 0;
}

A Banana

B Cherry

C Apple

D No output

💡 Hint

Check the first if condition and then the nested else-if.

The first if is false because ch is 'b', not 'a'. The else block contains another if that checks if ch == 'b', which is true, so it prints 'Banana'.

🔧 Debug

advanced

Identify the error in nested if-else code

What error will this code produce when compiled?

C

#include <stdio.h>

int main() {
    int x = 10;
    if (x > 5)
        if (x < 15)
            printf("Inside\n");
        else
            printf("Outside\n");
    return 0;
}

A Syntax error due to missing braces

B Runtime error

C No error, prints "Outside"

D No error, prints "Inside"

💡 Hint

Remember how else matches the nearest if without braces.

The else matches the inner if (x < 15). Since x=10, the condition is true, so it prints "Inside". No syntax or runtime error occurs.

🧠 Conceptual

advanced

Number of times inner block executes

Consider the code below. How many times will the inner printf("Hello\n") execute?

C

#include <stdio.h>

int main() {
    int i;
    for (i = 0; i < 5; i++) {
        if (i % 2 == 0) {
            if (i > 1) {
                printf("Hello\n");
            }
        }
    }
    return 0;
}

A 4

B 2

C 3

D 1

💡 Hint

Check values of i that are even and greater than 1.

i takes values 0,1,2,3,4. Even values are 0,2,4. Among these, i > 1 is true for 2 and 4. So the inner printf runs twice.

❓ Predict Output

expert

Output of nested if-else with multiple conditions

What is the output of this code snippet?

C

#include <stdio.h>

int main() {
    int a = 3, b = 7, c = 5;
    if (a > b) {
        if (b > c) {
            printf("X\n");
        } else {
            printf("Y\n");
        }
    } else {
        if (a > c) {
            printf("Z\n");
        } else {
            printf("W\n");
        }
    }
    return 0;
}

A W

B Z

C X

D Y

💡 Hint

Evaluate conditions stepwise: a > b, then b > c, then a > c.

a=3, b=7, c=5. a > b is false, so else block runs. Then a > c is false (3 > 5 is false), so it prints "W".

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+