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Embedded Cprogramming~10 mins

XOR for toggling bits in Embedded C - Interactive Code Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to toggle the 3rd bit of the variable flags.

Embedded C
flags = flags [1] (1 << 2);
Drag options to blanks, or click blank then click option'
A&
B|
C^
D<<
Attempts:
3 left
💡 Hint
Common Mistakes
Using & or | instead of ^ will not toggle bits correctly.
2fill in blank
medium

Complete the code to toggle the least significant bit of status.

Embedded C
status [1]= 1;
Drag options to blanks, or click blank then click option'
A^
B&
C|
D<<
Attempts:
3 left
💡 Hint
Common Mistakes
Using &= or |= will not toggle the bit.
3fill in blank
hard

Fix the error in toggling the 5th bit of control.

Embedded C
control = control [1] (1 << 4);
Drag options to blanks, or click blank then click option'
A^
B<<
C|
D&
Attempts:
3 left
💡 Hint
Common Mistakes
Using | sets the bit instead of toggling it.
4fill in blank
hard

Fill both blanks to toggle the 2nd bit and check if it was set before toggling.

Embedded C
if ((flags [1] (1 << 1)) != 0) {
    flags [2]= (1 << 1);
}
Drag options to blanks, or click blank then click option'
A&
B^
C|
D<<
Attempts:
3 left
💡 Hint
Common Mistakes
Using | to check bits is incorrect.
Using |= toggles incorrectly.
5fill in blank
hard

Fill all three blanks to toggle the 4th bit of reg only if it is currently set.

Embedded C
if (reg [1] (1 << 3)) {
    reg [2]= (1 << 3);
    result = reg [3] (1 << 3);
}
Drag options to blanks, or click blank then click option'
A&
B^
C|
D<<
Attempts:
3 left
💡 Hint
Common Mistakes
Using | instead of & to check or read bits.
Using |= instead of ^= to toggle bits.

Practice

(1/5)
1. What does the XOR operator (^) do when used to toggle bits in embedded C?
easy
A. It shifts bits to the left by the mask value.
B. It flips bits where the mask has 1s and leaves others unchanged.
C. It clears all bits to 0 regardless of the mask.
D. It sets all bits to 1 regardless of the mask.

Solution

  1. Step 1: Understand XOR operation basics

    XOR flips bits where the mask bit is 1 and keeps bits unchanged where mask bit is 0.
  2. Step 2: Apply to toggling bits

    Using XOR with a mask toggles only the bits set in the mask, leaving others unchanged.
  3. Final Answer:

    It flips bits where the mask has 1s and leaves others unchanged. -> Option B
  4. Quick Check:

    XOR toggles bits = D [OK]
Hint: XOR flips bits only where mask bits are 1 [OK]
Common Mistakes:
  • Thinking XOR sets bits to 1 always
  • Confusing XOR with AND or OR
  • Assuming XOR shifts bits
2. Which of the following is the correct syntax to toggle the 3rd bit (bit 2) of a variable flags using XOR in embedded C?
easy
A. flags = flags ^ (1 << 2);
B. flags = flags | (1 << 2);
C. flags = flags & ~(1 << 2);
D. flags = flags << 2;

Solution

  1. Step 1: Identify toggle operation syntax

    To toggle a bit, XOR (^) with a mask having 1 at that bit position is used.
  2. Step 2: Check each option

    flags = flags ^ (1 << 2); uses XOR with (1 << 2) which correctly toggles bit 2. Others set, clear, or shift bits incorrectly.
  3. Final Answer:

    flags = flags ^ (1 << 2); -> Option A
  4. Quick Check:

    Toggle bit syntax = B [OK]
Hint: Use XOR with (1 << bit_position) to toggle bits [OK]
Common Mistakes:
  • Using OR instead of XOR to toggle
  • Using AND with negated mask to clear bits
  • Using shift operator alone
3. What is the output of the following code snippet?
unsigned char flags = 0b00001100;
flags = flags ^ 0b00000110;
printf("0x%02X", flags);
medium
A. 0x06
B. 0x0E
C. 0x0A
D. 0x12

Solution

  1. Step 1: Convert initial flags and mask to binary

    flags = 00001100 (0x0C), mask = 00000110 (0x06)
  2. Step 2: Apply XOR operation bitwise

    00001100 ^ 00000110 = 00001010 (0x0A)
  3. Final Answer:

    0x0A -> Option C
  4. Quick Check:

    0x0C ^ 0x06 = 0x0A [OK]
Hint: XOR flips bits where mask is 1, calculate bitwise [OK]
Common Mistakes:
  • Adding values instead of XORing
  • Misreading binary bits
  • Confusing hex output format
4. The following code is intended to toggle bit 1 of status. What is the error?
unsigned int status = 0x01;
status = status & (1 << 1);
medium
A. The code should use OR operator to toggle bits.
B. The mask should be shifted by 2, not 1.
C. The variable should be unsigned char, not unsigned int.
D. Using AND instead of XOR will clear bits instead of toggling.

Solution

  1. Step 1: Analyze the operator used

    The code uses AND (&) with mask, which clears bits not set in mask instead of toggling.
  2. Step 2: Identify correct operator for toggling

    XOR (^) is the correct operator to toggle bits, not AND (&).
  3. Final Answer:

    Using AND instead of XOR will clear bits instead of toggling. -> Option D
  4. Quick Check:

    Toggle needs XOR, not AND [OK]
Hint: Toggle bits only with XOR, AND clears bits [OK]
Common Mistakes:
  • Confusing AND with XOR for toggling
  • Incorrect bit shift amount
  • Wrong data type assumption
5. You have a 8-bit register reg initially set to 0b10101010. You want to toggle bits 1, 3, and 5 simultaneously using XOR. Which mask should you use and what will be the final value of reg after toggling?
hard
A. Mask: 0b00101010; Final reg: 0b10000000
B. Mask: 0b01010101; Final reg: 0b11111111
C. Mask: 0b00101010; Final reg: 0b10000010
D. Mask: 0b00010101; Final reg: 0b10111111

Solution

  1. Step 1: Create mask for bits 1, 3, and 5

    Bits are zero-indexed from right. Bit 1 = 0b00000010, bit 3 = 0b00001000, bit 5 = 0b00100000. OR them: 0b00101010.
  2. Step 2: XOR reg with mask

    reg = 0b10101010 ^ 0b00101010 = 0b10000000.
  3. Final Answer:

    Mask: 0b00101010; Final reg: 0b10000000 -> Option A
  4. Quick Check:

    Toggle bits 1,3,5 with mask 0x2A = 0x80 [OK]
Hint: Combine bits with OR for mask, XOR to toggle all at once [OK]
Common Mistakes:
  • Wrong mask bits set
  • Miscounting bit positions
  • Incorrect XOR result calculation
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