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.

Practice

1. What does the expression (num & (1 << 3)) != 0 check in embedded C?

easy

A. If the 1st bit (bit 0) of num is set to 1

B. If the 4th bit (bit 3) of num is set to 1

C. If num is equal to 8

D. If num is an even number

Solution

1. Step 1: Understand the bit shift operation

   The expression 1 << 3 shifts 1 left by 3 bits, creating a mask with only the 4th bit set (value 8).

2. Step 2: Apply bitwise AND to check the bit

   The bitwise AND num & (1 << 3) isolates the 4th bit of num. If this result is not zero, the bit is set.

3. Final Answer:

   Checks if the 4th bit (bit 3) of num is set to 1 -> Option B

4. Quick Check:

   Bitwise AND with mask = non-zero means bit set [OK]

Hint: Use (1 << bit) & num to check bit set quickly [OK]

Common Mistakes:

• Confusing bit position with bit value
• Forgetting bits start at 0
• Using OR instead of AND to check bits

2. Which of the following is the correct syntax to check if the 2nd bit (bit 1) of variable x is set in embedded C?

easy

A. if (x & 1) { /* bit is set */ }

B. if (x | (1 << 1)) { /* bit is set */ }

C. if (x << 1) { /* bit is set */ }

D. if (x & (1 << 1)) { /* bit is set */ }

Solution

1. Step 1: Identify the mask for bit 1

   The mask for the 2nd bit is 1 << 1, which equals 2.

2. Step 2: Use bitwise AND to check the bit

   Using x & (1 << 1) checks if bit 1 is set. If the result is non-zero, the bit is set.

3. Final Answer:

   if (x & (1 << 1)) { /* bit is set */ } -> Option D

4. Quick Check:

   Bitwise AND with shifted 1 checks bit [OK]

Hint: Use & with (1 << bit) to check bits, not | or << alone [OK]

Common Mistakes:

• Using bitwise OR (|) instead of AND (&)
• Using left shift on variable instead of mask
• Checking wrong bit position

3. What is the output of the following code snippet?

unsigned int num = 13; // binary 1101
if (num & (1 << 2))
    printf("Bit 2 is set\n");
else
    printf("Bit 2 is not set\n");

medium

A. Compilation error

B. Bit 2 is not set

C. Bit 2 is set

D. No output

Solution

1. Step 1: Convert num to binary and identify bit 2

   Number 13 in binary is 1101. Bits are numbered from 0 (rightmost). Bit 2 is the third bit from right, which is 1.

2. Step 2: Check bit 2 using bitwise AND

   The expression num & (1 << 2) equals 13 & 4 which is 4 (non-zero), so the condition is true.

3. Final Answer:

   Bit 2 is set -> Option C

4. Quick Check:

   Bit 2 in 1101 is 1, so output is "Bit 2 is set" [OK]

Hint: Convert number to binary, check bit position with mask [OK]

Common Mistakes:

• Miscounting bit positions starting at 1 instead of 0
• Assuming bit 2 is zero without checking
• Confusing bitwise AND with OR

4. The following code is intended to check if bit 5 of flags is set. What is the error?

unsigned int flags = 0x20;
if (flags & 5 << 1) {
    // bit is set
}

medium

A. Incorrect operator precedence causes wrong mask

B. Bit shift should be on variable, not constant

C. Missing parentheses around bit shift

D. Using bitwise OR instead of AND

Solution

1. Step 1: Analyze operator precedence in expression

   The expression flags & 5 << 1 is evaluated as flags & (5 << 1) due to operator precedence (<< higher than &).

2. Step 2: Understand the intended mask and fix

   The intended mask is 1 << 5 to check bit 5. The current expression does not create the correct mask.

3. Final Answer:

   Incorrect operator precedence causes wrong mask -> Option A

4. Quick Check:

   Use parentheses: flags & (1 << 5) to fix [OK]

Hint: Always use parentheses around (1 << bit) mask [OK]

Common Mistakes:

• Not using parentheses causing wrong evaluation
• Shifting wrong constant (5 instead of 1)
• Confusing bit positions and mask values

5. You have an 8-bit status register status. Write a condition to check if either bit 1 or bit 3 is set. Which of the following is correct?

hard

A. if (status & ((1 << 1) | (1 << 3))) { /* bit 1 or 3 set */ }

B. if ((status & (1 << 1)) && (status & (1 << 3))) { /* bit 1 or 3 set */ }

C. if (status | (1 << 1) | (1 << 3)) { /* bit 1 or 3 set */ }

D. if (status & (1 << (1 + 3))) { /* bit 1 or 3 set */ }

Solution

1. Step 1: Create mask for bits 1 and 3

   Use bitwise OR to combine masks: (1 << 1) | (1 << 3) creates a mask with bits 1 and 3 set.

2. Step 2: Use bitwise AND to check if either bit is set

   Check if status & mask is non-zero to confirm if either bit is set.

3. Final Answer:

   if (status & ((1 << 1) | (1 << 3))) { /* bit 1 or 3 set */ } -> Option A

4. Quick Check:

   Combine masks with |, check with & for any bit set [OK]

Hint: Combine bit masks with | inside parentheses before & [OK]

Common Mistakes:

• Using && instead of | to combine masks
• Using + instead of | causing wrong mask
• Using bitwise OR on variable instead of mask