Checking if a bit is set in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this bit check code?

Consider the following C code snippet that checks if the 3rd bit (bit index 2) of a number is set. What will be printed?

Embedded C

unsigned int num = 10; // binary 1010
if (num & (1 << 2)) {
    printf("Bit is set\n");
} else {
    printf("Bit is not set\n");
}

A Bit is not set

B Bit is set

C Compilation error

D Runtime error

💡 Hint

Remember that bits are counted from 0 starting at the least significant bit.

The number 10 in binary is 1010. The bits are indexed 3 2 1 0 from left to right. Bit 2 (third bit) is 0, so the condition is false and "Bit is not set" is printed.

❓ Predict Output

intermediate

What does this code print when checking bit 0?

Given the code below, what will be the output?

Embedded C

unsigned char val = 1; // binary 00000001
if (val & 1) {
    printf("LSB is set\n");
} else {
    printf("LSB is not set\n");
}

A Compilation error

B LSB is not set

C LSB is set

D Runtime error

💡 Hint

LSB means least significant bit, which is bit 0.

The value 1 in binary is 00000001, so bit 0 is set. The condition is true and "LSB is set" is printed.

❓ Predict Output

advanced

Output of checking bit 5 in a signed integer

What will this code print?

Embedded C

int x = -32; // binary representation depends on two's complement
if (x & (1 << 5)) {
    printf("Bit 5 is set\n");
} else {
    printf("Bit 5 is not set\n");
}

A Bit 5 is not set

B Bit 5 is set

C Undefined behavior

D Compilation error

💡 Hint

Remember two's complement representation for negative numbers.

In two's complement, -32 is represented with bit 5 set (since 32 = 1 << 5). So the bitwise AND is non-zero and "Bit 5 is set" is printed.

❓ Predict Output

advanced

What error does this code produce?

What error will this code cause?

Embedded C

unsigned int num = 5;
if (num & (1U << 32)) {
    printf("Bit 32 is set\n");
} else {
    printf("Bit 32 is not set\n");
}

A Undefined behavior due to shift overflow

B No error, prints "Bit 32 is not set"

C Compilation error

D Runtime error

💡 Hint

Check the size of unsigned int and the shift operation limits.

Shifting 1 by 32 bits on a 32-bit unsigned int is undefined behavior in C. This can cause unpredictable results or runtime errors.

🧠 Conceptual

expert

How many bits are set in this mask?

Given the mask 0xAA55 (hexadecimal), how many bits are set to 1 in its 16-bit binary representation?

A 12

B 6

C 10

D 8

💡 Hint

Convert hex to binary and count the 1s.

0xAA55 in binary is 1010101001010101. Counting the 1s gives 8 bits set.