Checking if a bit is set in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the 3rd bit (bit 2) of variable 'flags' is set.

Embedded C

if (flags & [1]) {
    // bit is set
}

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

A1 << 1

B1 << 3

C2 << 1

D1 << 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong bit position shift.

Confusing bit numbering starting from 1 instead of 0.

To check the 3rd bit (bit 2), we use 1 shifted left by 2 positions: 1 << 2.

2fill in blank

medium

Complete the code to check if the 5th bit (bit 4) of 'status' is set.

Embedded C

if ((status & [1]) != 0) {
    // bit is set
}

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

A1 << 4

B4 << 1

C1 << 5

D1 << 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 << 5 which checks the 6th bit.

Not comparing the result to zero.

The 5th bit is bit 4 (counting from 0), so use 1 << 4.

3fill in blank

hard

Fix the error in the code to correctly check if bit 1 of 'reg' is set.

Embedded C

if (reg & [1]) {
    // bit is set
}

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

A1 << 2

B1 << 1

C2 << 2

D1 << 0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 << 2 which checks bit 2 instead of bit 1.

Using 1 << 0 which checks bit 0.

Bit 1 means shifting 1 by 1: 1 << 1.

4fill in blank

hard

Fill both blanks to complete the for loop that maps each bit position to whether it is set in 'value'.

Embedded C

for (int [1] = 0; [1] < 8; [1]++) {
    bits[[1]] = (value & (1 << [2])) != 0;
}

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

Ai

Bj

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for loop and bit shift causing errors.

Using undefined variables.

Use the same loop variable i for both the loop and the bit shift.

5fill in blank

hard

Fill all three blanks to create a function that returns 1 if the bit at 'pos' in 'num' is set, else 0.

Embedded C

int is_bit_set(unsigned int num, int [1]) {
    return (num & (1 << [2])) [3] 0 ? 1 : 0;
}

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

Apos

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for bit position.

Using '==' which returns 1 if bit is not set.

The function uses pos as parameter and bit position, and checks if the bit is not zero.