Left shift and right shift in C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of left shift operation

What is the output of this C code snippet?

C

int main() {
    unsigned int x = 5; // binary: 0000 0101
    unsigned int y = x << 2;
    printf("%u", y);
    return 0;
}

A 40

B 10

C 5

D 20

💡 Hint

Left shift by 2 moves bits two places to the left, multiplying by 4.

5 in binary is 00000101. Shifting left by 2 bits gives 00010100, which is 20 in decimal.

❓ Predict Output

intermediate

Output of right shift operation

What is the output of this C code snippet?

C

int main() {
    unsigned int x = 32; // binary: 0010 0000
    unsigned int y = x >> 3;
    printf("%u", y);
    return 0;
}

A 4

B 8

C 16

D 64

💡 Hint

Right shift by 3 moves bits three places to the right, dividing by 8.

32 in binary is 00100000. Shifting right by 3 bits gives 00000100, which is 4 in decimal.

🧠 Conceptual

advanced

Effect of left shift on signed integers

Consider the following code. What is the output and why?

C

int main() {
    int x = 1073741824; // 2^30
    int y = x << 2;
    printf("%d", y);
    return 0;
}

A -1073741824

B 1073741824

C 0

D 4294967296

💡 Hint

Left shift by 2 shifts bit 30 to bit 32, which is discarded in 32-bit int.

2^30 (0x40000000, bit 30 set) shifted left by 2 moves the bit to position 32 (discarded), resulting in a negative number due to overflow (two's complement).

🔧 Debug

advanced

Identify the error in right shift usage

What error will this code produce when compiled and run?

C

int main() {
    int x = -16;
    unsigned int y = x >> 2;
    printf("%u", y);
    return 0;
}

A Prints -4

B Compilation error due to mixing signed and unsigned types

C Prints a large positive number due to sign extension and conversion

D Runtime error: undefined behavior

💡 Hint

Right shifting a negative signed int keeps the sign bit, then converting to unsigned prints a large number.

Right shifting negative signed int keeps sign bits (arithmetic shift). Assigning to unsigned int prints the large unsigned value of that bit pattern.

🚀 Application

expert

Calculate the number of bits set after shifts

Given the code below, what is the number of bits set to 1 in variable z after execution?

C

int main() {
    unsigned int x = 0b1011; // decimal 11
    unsigned int y = x << 3;  // shift left by 3
    unsigned int z = y >> 2;  // shift right by 2
    int count = 0;
    for (int i = 0; i < 32; i++) {
        if ((z & (1 << i)) != 0) count++;
    }
    printf("%d", count);
    return 0;
}

A 2

B 3

C 4

D 1

💡 Hint

Track the bits step by step after each shift.

x = 0b1011 (bits set: 3). Left shift by 3: 0b1011000 (bits set: 3). Right shift by 2: 0b10110 (bits set: 3). So count is 3.