Clearing a specific bit in a register in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to clear bit 3 of the variable reg.

Embedded C

reg = reg & [1];

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

A1 << 3

B1 >> 3

C~(1 >> 3)

D~(1 << 3)

Attempts:

3 left

💡 Hint

Common Mistakes

Using bitwise OR instead of AND.

Not inverting the mask before AND.

Shifting bits in the wrong direction.

To clear bit 3, we use bitwise AND with the complement of (1 shifted left by 3).

2fill in blank

medium

Complete the code to clear the bit specified by bit_pos in reg.

Embedded C

reg &= [1];

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

A~(1 << bit_pos)

B~(1 >> bit_pos)

C1 >> bit_pos

D1 << bit_pos

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to invert the mask.

Using OR instead of AND.

Shifting bits in the wrong direction.

To clear a bit at position bit_pos, AND with the complement of (1 shifted left by bit_pos).

3fill in blank

hard

Fix the error in the code that attempts to clear bit 2 of reg.

Embedded C

reg = reg & [1];

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

A1 << 2

B~(1 << 2)

C~(1 >> 2)

D1 >> 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using a mask without inversion.

Shifting bits right instead of left.

Using OR instead of AND.

The mask must be the complement of (1 shifted left by 2) to clear bit 2 correctly.

4fill in blank

hard

Fill both blanks to clear bit 5 in reg using a mask.

Embedded C

unsigned int mask = [1];
reg = reg & [2];

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

A~(1 << 5)

B1 << 5

Cmask

D~mask

Attempts:

3 left

💡 Hint

Common Mistakes

Not inverting the mask before AND.

Using the mask directly without inversion.

Shifting bits in the wrong direction.

Create the mask by shifting 1 left by 5, then clear bit 5 by ANDing reg with the complement of the mask.

5fill in blank

hard

Fill all three blanks to clear the bit at position pos in reg using a mask variable.

Embedded C

unsigned int mask = [1];
unsigned int inv_mask = [2];
reg = reg [3] inv_mask;

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

A1 << pos

B~mask

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using OR instead of AND to clear the bit.

Not inverting the mask before AND.

Shifting bits right instead of left.

Create the mask by shifting 1 left by pos, invert it to get inv_mask, then clear the bit by ANDing reg with inv_mask.

Practice

1. What does the expression register &= ~(1 << 3); do in embedded C?

easy

A. Sets the 3rd bit of the register to 1

B. Clears the 3rd bit of the register (sets it to 0) without changing other bits

C. Toggles the 3rd bit of the register

D. Clears all bits of the register

Solution

1. Step 1: Understand bit shifting and negation

   The expression 1 << 3 shifts 1 to the left by 3 bits, creating a mask with only the 3rd bit set (00001000).

2. Step 2: Apply bitwise NOT and AND assignment

   The bitwise NOT ~(1 << 3) flips all bits, making the 3rd bit 0 and others 1. ANDing this with the register clears only the 3rd bit.

3. Final Answer:

   Clears the 3rd bit of the register (sets it to 0) without changing other bits -> Option B

4. Quick Check:

   Clear bit 3 = register &= ~(1 << 3) [OK]

Hint: Use AND with NOT mask to clear a bit [OK]

Common Mistakes:

• Confusing clearing with setting a bit
• Using OR instead of AND
• Not applying bitwise NOT before AND

2. Which of the following is the correct syntax to clear bit 5 of a variable named reg?

easy

A. reg &= ~(1 << 5);

B. reg |= ~(1 << 5);

C. reg = reg & (1 << 5);

D. reg = reg | (1 << 5);

Solution

1. Step 1: Identify the operation to clear a bit

   Clearing a bit means setting it to 0. This is done by ANDing with the negation of the bit mask.

2. Step 2: Check each option's operation

   reg &= ~(1 << 5); uses &= ~(1 << 5), which correctly clears bit 5. Others either set or incorrectly manipulate bits.

3. Final Answer:

   reg &= ~(1 << 5); -> Option A

4. Quick Check:

   Clear bit 5 = reg &= ~(1 << 5) [OK]

Hint: Clear bit with AND and NOT mask [OK]

Common Mistakes:

• Using OR (|) instead of AND (&)
• Forgetting to negate the mask
• Assigning instead of using compound assignment

3. Given unsigned char reg = 0b11111111; and the operation reg &= ~(1 << 2);, what is the binary value of reg after execution?

medium

A. 0b11111011

B. 0b11111101

C. 0b11110111

D. 0b11111110

Solution

1. Step 1: Identify the bit to clear

   Bit 2 means the third bit from the right (0-based index). The mask is 1 << 2 = 0b00000100.

2. Step 2: Apply clearing operation

   Negate mask: ~0b00000100 = 0b11111011. AND with reg: 0b11111111 & 0b11111011 = 0b11111011.

3. Final Answer:

   0b11111011 -> Option A

4. Quick Check:

   Clear bit 2 from 0xFF = 0xFB (binary 11111011) [OK]

Hint: Clear bit 2 flips that bit to 0 only [OK]

Common Mistakes:

• Clearing wrong bit index
• Not negating the mask
• Confusing bit positions (1-based vs 0-based)

4. What is wrong with this code snippet if the goal is to clear bit 4 of reg?

reg = reg & (1 << 4);

medium

A. It causes a syntax error

B. It correctly clears bit 4

C. It sets all bits except bit 4 to 0, not clearing bit 4

D. It toggles bit 4 instead of clearing

Solution

1. Step 1: Analyze the operation performed

   The expression reg & (1 << 4) keeps only bit 4 and clears all others, effectively zeroing all bits except bit 4.

2. Step 2: Compare with correct clearing method

   To clear bit 4, we must AND with the negation of the mask: reg &= ~(1 << 4);. The given code does not do this.

3. Final Answer:

   It sets all bits except bit 4 to 0, not clearing bit 4 -> Option C

4. Quick Check:

   AND with mask keeps only that bit [OK]

Hint: AND with mask keeps only that bit, not clears it [OK]

Common Mistakes:

• Using AND with mask instead of negated mask
• Confusing clearing with masking
• Expecting toggling behavior

5. You have a 16-bit register uint16_t reg = 0xFFFF;. You want to clear bits 0, 4, and 15 simultaneously. Which code correctly does this?

hard

A. reg |= ( (1 << 0) | (1 << 4) | (1 << 15) );

B. reg &= ( (1 << 0) | (1 << 4) | (1 << 15) );

C. reg |= ~( (1 << 0) | (1 << 4) | (1 << 15) );

D. reg &= ~( (1 << 0) | (1 << 4) | (1 << 15) );

Solution

1. Step 1: Create a mask for bits to clear

   Bits 0, 4, and 15 correspond to masks 1<<0, 1<<4, and 1<<15. OR them to combine: (1<<0) | (1<<4) | (1<<15).

2. Step 2: Negate mask and AND with register

   Negate the combined mask to get zeros at bits 0,4,15 and ones elsewhere. AND with reg clears those bits.

3. Final Answer:

   reg &= ~( (1 << 0) | (1 << 4) | (1 << 15) ); -> Option D

4. Quick Check:

   Clear multiple bits by AND with negated OR mask [OK]

Hint: Combine bits with OR, negate, then AND to clear multiple bits [OK]

Common Mistakes:

• Using OR instead of AND to clear bits
• Not negating the combined mask
• Clearing only one bit instead of all three