Register bit manipulation patterns in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set bit 3 of the register.

Embedded C

REG |= (1 << [1]);

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

A3

B2

C4

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong bit number.

Forgetting to use parentheses around (1 << bit).

Bit 3 is set by shifting 1 left by 3 positions.

2fill in blank

medium

Complete the code to clear bit 5 of the register.

Embedded C

REG &= ~(1 << [1]);

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

A2

B3

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of 1 for the mask.

Not negating the mask before AND operation.

To clear bit 5, shift 1 left by 5 and invert it with ~.

3fill in blank

hard

Fix the error in the code to toggle bit 2 of the register.

Embedded C

REG [1]= (1 << 2);

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

A|

B^

C&

D~

Attempts:

3 left

💡 Hint

Common Mistakes

Using |= which sets the bit instead of toggling.

Using &= which clears bits.

Toggle a bit using XOR (^=) with the bit mask.

4fill in blank

hard

Fill both blanks to check if bit 4 is set in the register.

Embedded C

if ((REG [1] (1 [2] 4)) != 0) {
    // bit 4 is set
}

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

A&

B|

C<<

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using | instead of & which always returns non-zero.

Using right shift >> instead of left shift <<.

Use bitwise AND (&) with 1 shifted left (<<) by 4 to check bit 4.

5fill in blank

hard

Fill all three blanks to set bits 0, 2, and 7 in the register.

Embedded C

REG |= ([1] << 0) | ([2] << 2) | ([3] << 7);

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

A1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of 1 for the bit mask.

Not combining masks with OR operator.

Set bits by shifting 1 to each bit position and OR-ing them together.

Practice

1. Which bitwise operator is used to set a specific bit in a register without changing other bits?
Example: Set bit 3 in register REG

easy

A. Bitwise OR (|)

B. Bitwise AND (&)

C. Bitwise XOR (^)

D. Bitwise NOT (~)

Solution

1. Step 1: Understand setting a bit

   Setting a bit means making that bit 1 without changing others.

2. Step 2: Use bitwise OR

   Bitwise OR with 1 at the target bit sets it to 1, others stay same.

3. Final Answer:

   Bitwise OR (|) -> Option A

4. Quick Check:

   Set bit = OR [OK]

Hint: Use OR (|) to set bits without changing others [OK]

Common Mistakes:

• Using AND (&) to set bits (it clears bits)
• Using XOR (^) which toggles bits
• Using NOT (~) which inverts all bits

2. Which of the following C code snippets correctly clears bit 2 of a register named REG?

easy

A. REG = REG | (1 << 2);

B. REG = REG << 2;

C. REG = REG ^ (1 << 2);

D. REG = REG & ~(1 << 2);

Solution

1. Step 1: Understand clearing a bit

   Clearing means setting the bit to 0 without changing others.

2. Step 2: Use AND with NOT mask

   AND with NOT of (1 shifted to bit 2) clears bit 2.

3. Final Answer:

   REG = REG & ~(1 << 2); -> Option D

4. Quick Check:

   Clear bit = AND with NOT mask [OK]

Hint: Clear bits with AND and NOT mask (~) [OK]

Common Mistakes:

• Using OR (|) which sets bits
• Using XOR (^) which toggles bits
• Using shift (<<) alone which moves bits

3. Given REG = 0b00001100;, what is the value of REG after executing REG ^= (1 << 3);?

medium

A. 0b00001100

B. 0b00000100

C. 0b00011100

D. 0b00001000

Solution

1. Step 1: Understand XOR toggling

   XOR toggles the bit: if 1 becomes 0, if 0 becomes 1.

2. Step 2: Toggle bit 3 of REG

   Bit 3 is currently 1 (0b00001100). Toggling it changes it to 0, so REG becomes 0b00000100.

3. Final Answer:

   0b00000100 -> Option B

4. Quick Check:

   XOR toggles bit 3 from 1 to 0 [OK]

Hint: XOR (^) flips bits: 1->0 or 0->1 [OK]

Common Mistakes:

• Thinking XOR sets bit instead of toggling
• Confusing bit positions (count from 0)
• Not converting binary correctly

4. What is wrong with this code to check if bit 4 of REG is set?

if (REG & (1 << 4) == 1) {
  // bit 4 is set
}

medium

A. The comparison '== 1' is incorrect; it should check if result is non-zero.

B. The shift operator '<<' is used wrongly; it should be '>>'.

C. The bit number 4 is out of range for an 8-bit register.

D. The code should use OR (|) instead of AND (&) to check bits.

Solution

1. Step 1: Understand bit check expression

   REG & (1 << 4) returns non-zero if bit 4 is set, zero if not.

2. Step 2: Fix comparison

   Comparing to 1 is wrong because bit 4 corresponds to value 16, not 1. Should check if result != 0.

3. Final Answer:

   The comparison '== 1' is incorrect; it should check if result is non-zero. -> Option A

4. Quick Check:

   Check bits by testing non-zero result [OK]

Hint: Check bits by testing if (REG & mask) != 0 [OK]

Common Mistakes:

• Comparing bit check result to 1 instead of non-zero
• Using shift right (>>) instead of left (<<)
• Using OR (|) to check bits which sets bits

5. You want to toggle bits 0, 2, and 5 of an 8-bit register REG simultaneously. Which code correctly does this?

hard

A. REG = REG & ~( (1 << 0) | (1 << 2) | (1 << 5) );

B. REG = REG | ( (1 << 0) & (1 << 2) & (1 << 5) );

C. REG ^= (1 << 0) | (1 << 2) | (1 << 5);

D. REG ^= (1 << 1) | (1 << 3) | (1 << 5);

Solution

1. Step 1: Understand toggling multiple bits

   Toggling means flipping bits. XOR (^) with mask flips bits where mask has 1.

2. Step 2: Create mask for bits 0, 2, and 5

   Mask is (1<<0) | (1<<2) | (1<<5) which sets bits 0, 2, and 5 to 1.

3. Step 3: Apply XOR with mask

   REG ^= mask flips those bits simultaneously.

4. Final Answer:

   REG ^= (1 << 0) | (1 << 2) | (1 << 5); -> Option C

5. Quick Check:

   Toggle bits with XOR and combined mask [OK]

Hint: Toggle multiple bits with XOR and combined mask [OK]

Common Mistakes:

• Using AND with NOT mask which clears bits
• Using AND (&) instead of OR (|) to combine masks
• Using wrong bit positions in mask