OR for setting bits in Embedded C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the OR operator (|) do when used to set bits in embedded C?

The OR operator sets specific bits to 1 without changing other bits. It combines bits so if either bit is 1, the result bit is 1.

beginner

How do you set the 3rd bit of a variable 'flags' using OR in embedded C?

Use: flags = flags | (1 << 2); This shifts 1 to the 3rd bit position and ORs it with flags to set that bit.

beginner

Why is OR used instead of AND to set bits?

AND clears bits (sets to 0) if the mask bit is 0. OR sets bits to 1 if the mask bit is 1, which is what we want to set bits.

beginner

What happens if you OR a variable with 0?

The variable stays the same because OR with 0 leaves bits unchanged.

beginner

Explain the expression: flags |= (1 << 5);

It sets the 6th bit of 'flags' to 1. The |= operator is shorthand for flags = flags | (1 << 5).

What does the expression 'flags | (1 << 3)' do?

A Sets the 4th bit of flags to 1

B Clears the 4th bit of flags

C Toggles the 4th bit of flags

D Checks if the 4th bit of flags is 1

Shifting 1 by 3 moves it to the 4th bit position (counting from 0). OR sets that bit to 1.

Which operator is used to set bits without affecting other bits?

A XOR (^)

B AND (&)

C OR (|)

D NOT (~)

OR sets bits to 1 where the mask has 1s, leaving other bits unchanged.

What is the result of 'flags | 0'?

A flags unchanged

B All bits set to 1

C All bits cleared to 0

D Error

OR with 0 leaves all bits unchanged.

How do you set the 1st bit of a variable 'x' using OR?

A x = x << 1;

B x = x & 1;

C x = x ^ 1;

D x = x | 1;

OR with 1 sets the least significant bit (1st bit) to 1.

What does 'flags |= (1 << n);' do?

A Clears the nth bit

B Sets the nth bit

C Toggles the nth bit

D Checks the nth bit

It sets the nth bit of flags to 1 using OR and compound assignment.

Explain how the OR operator is used to set specific bits in a variable in embedded C.

Think about how 1 shifted left creates a mask to set one bit.

Describe the difference between using OR and AND operators when manipulating bits.

Consider what happens when you combine bits with 1 or 0.

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