OR for setting bits in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set bit 2 of the variable 'flags'.

Embedded C

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 '&=' clears bits instead of setting them.

Using '^=' toggles bits instead of just setting them.

Using the OR assignment operator '|=' sets bit 2 without changing other bits.

2fill in blank

medium

Complete the code to set bits 0 and 3 of 'status' using OR.

Embedded C

status [1] (1 << 0) | (1 << 3);

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

A&=

B=

C^=

D|=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' overwrites all bits, losing previous data.

Using '&=' clears bits instead of setting them.

The '|=' operator sets bits 0 and 3 without changing other bits.

3fill in blank

hard

Fix the error in the code to correctly set bit 5 of 'control'.

Embedded C

control [1] 1 << 5;

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

A=

B|=

C^=

D&=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' overwrites all bits, losing previous settings.

Using '&=' clears bits instead of setting them.

Using '|=' with '1 << 5' sets bit 5 correctly.

4fill in blank

hard

Fill both blanks to set bits 1 and 4 of 'reg' using OR.

Embedded C

reg [1] (1 << 1 [2] 1 << 4);

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

A|=

B&

C|

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&' instead of '|' to combine bits clears bits.

Using '^' toggles bits instead of setting them.

Use '|=' to assign and '|' to combine bits 1 and 4.

5fill in blank

hard

Fill all three blanks to set bits 0, 2, and 7 of 'flags' using OR.

Embedded C

flags [1] (1 [2] 1 << 2 [3] 1 << 7);

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

A|=

B&

C|

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&' clears bits instead of setting them.

Using '^' toggles bits instead of setting them.

Use '|=' to assign and '|' to combine multiple bits to set bits 0, 2, and 7.

Practice

1. What does the OR operation do when used to set bits in a variable in embedded C?

easy

A. It toggles specified bits from 0 to 1 or 1 to 0.

B. It clears specified bits to 0 without changing other bits.

C. It sets specified bits to 1 without changing other bits.

D. It shifts bits to the left by one position.

Solution

1. Step 1: Understand OR operation effect on bits

   The OR operation sets bits to 1 if either bit is 1, so it can set bits without clearing others.

2. Step 2: Apply to setting bits

   Using OR with a mask having 1s in desired positions sets those bits to 1, leaving others unchanged.

3. Final Answer:

   It sets specified bits to 1 without changing other bits. -> Option C

4. Quick Check:

   OR sets bits to 1 without clearing others [OK]

Hint: OR with mask sets bits to 1, leaves others unchanged [OK]

Common Mistakes:

• Confusing OR with AND operation
• Thinking OR clears bits instead of setting
• Assuming OR toggles bits
• Mixing OR with bit shifting

2. Which of the following is the correct syntax to set bits using OR in embedded C?

easy

A. variable |= mask;

B. variable = variable & mask;

C. variable = variable ^ mask;

D. variable &= ~mask;

Solution

1. Step 1: Identify syntax for setting bits with OR

   The operator '|=' is the shorthand for 'variable = variable | mask;', which sets bits.

2. Step 2: Check other options

   variable = variable & mask; uses AND, which clears bits; variable = variable ^ mask; uses XOR, which toggles bits; variable &= ~mask; clears bits using AND with negated mask.

3. Final Answer:

   variable |= mask; -> Option A

4. Quick Check:

   Use '|=' to set bits quickly [OK]

Hint: Use '|=' operator to set bits easily [OK]

Common Mistakes:

• Using '&' instead of '|' to set bits
• Confusing XOR (^) with OR (|)
• Forgetting shorthand '|=' syntax
• Using negation incorrectly

3. What is the value of flags after executing the following code?

unsigned char flags = 0x12;  // binary 00010010
unsigned char mask = 0x05;   // binary 00000101
flags = flags | mask;

medium

A. 0x05 (binary 00000101)

B. 0x17 (binary 00010111)

C. 0x10 (binary 00010000)

D. 0x07 (binary 00000111)

Solution

1. Step 1: Convert values to binary and apply OR

   flags = 00010010, mask = 00000101. OR operation sets bits where either is 1.

2. Step 2: Perform bitwise OR

   00010010 OR 00000101 = 00010111 (binary), which is 0x17 in hex.

3. Final Answer:

   0x17 (binary 00010111) -> Option B

4. Quick Check:

   0x12 | 0x05 = 0x17 [OK]

Hint: OR bits: 1 if either bit is 1 [OK]

Common Mistakes:

• Adding values instead of ORing
• Confusing OR with AND
• Misreading binary values
• Forgetting to convert hex to binary

4. Identify the error in the following code snippet that tries to set bits using OR:

unsigned int reg = 0x0F;
unsigned int mask = 0x30;
reg = reg & mask;

medium

A. Using '&' instead of '|' to set bits

B. Mask value is incorrect

C. Variable types should be 'char' not 'int'

D. Missing semicolon after mask declaration

Solution

1. Step 1: Understand the intended operation

   The code aims to set bits using OR, but uses '&' which clears bits instead.

2. Step 2: Identify correct operator

   To set bits, use '|' operator, so 'reg = reg | mask;' or 'reg |= mask;' is correct.

3. Final Answer:

   Using '&' instead of '|' to set bits -> Option A

4. Quick Check:

   Use '|' to set bits, '&' clears bits [OK]

Hint: Use '|' not '&' to set bits [OK]

Common Mistakes:

• Confusing AND with OR
• Assuming '&' sets bits
• Ignoring operator precedence
• Not testing code output

5. You have a 16-bit register reg initially set to 0x0000. You want to set bits 0, 3, and 15 to 1 without changing other bits. Which code correctly does this?

hard

A. reg = reg ^ ((1 << 0) | (1 << 3) | (1 << 15));

B. reg = reg & ((1 << 0) | (1 << 3) | (1 << 15));

C. reg |= (1 << 1) | (1 << 2) | (1 << 14);

D. reg |= (1 << 0) | (1 << 3) | (1 << 15);

Solution

1. Step 1: Identify bits to set and create mask

   Bits 0, 3, and 15 correspond to (1<<0), (1<<3), and (1<<15). OR them to create mask.

2. Step 2: Use OR assignment to set bits without changing others

   Use 'reg |= mask;' to set bits. reg |= (1 << 0) | (1 << 3) | (1 << 15); matches this.

3. Step 3: Check other options

   reg = reg & ((1 << 0) | (1 << 3) | (1 << 15)); uses AND which clears bits; reg |= (1 << 1) | (1 << 2) | (1 << 14); sets wrong bits; reg = reg ^ ((1 << 0) | (1 << 3) | (1 << 15)); toggles bits which may not be desired.

4. Final Answer:

   reg |= (1 << 0) | (1 << 3) | (1 << 15); -> Option D

5. Quick Check:

   Use OR with mask of bits to set [OK]

Hint: Use '|=' with bit-shifted mask for setting bits [OK]

Common Mistakes:

• Using '&' instead of '|' to set bits
• Setting wrong bit positions
• Using XOR which toggles bits
• Not combining bits into one mask