Setting a specific bit in a register in Embedded C

Introduction

We set a specific bit in a register to turn on a feature or control hardware without changing other settings.

Turning on an LED connected to a microcontroller pin.

Enabling a sensor by setting its control bit.

Configuring communication settings by setting bits in a control register.

Activating interrupts by setting the interrupt enable bit.

Changing device modes by setting mode bits in a register.

Syntax

Embedded C

REGISTER |= (1 << BIT_POSITION);

REGISTER is the name of the hardware register.

BIT_POSITION is the zero-based position of the bit to set (0 for least significant bit).

Examples

Sets bit 3 of PORTA register to 1, turning on the connected pin.

Embedded C

PORTA |= (1 << 3);

Sets bit 0 of CONTROL_REG to enable a feature.

Embedded C

CONTROL_REG |= (1 << 0);

Sets the highest bit (bit 7) of STATUS register.

Embedded C

STATUS |= (1 << 7);

Sample Program

This program starts with all bits off in registerA. It sets bit 2 to 1 using the bitwise OR and shift. Then it prints the register bits from left to right.

Embedded C

#include <stdio.h>

int main() {
    unsigned char registerA = 0b00000000; // all bits off
    int bit_to_set = 2; // we want to set bit 2

    // Set bit 2
    registerA |= (1 << bit_to_set);

    printf("Register after setting bit %d: 0b", bit_to_set);
    for (int i = 7; i >= 0; i--) {
        printf("%d", (registerA >> i) & 1);
    }
    printf("\n");

    return 0;
}

Important Notes

Using |= keeps other bits unchanged while setting the chosen bit.

Bit positions start at 0 from the right (least significant bit).

Make sure the register variable is wide enough to hold the bits you want to set.

Summary

Use REGISTER |= (1 << BIT_POSITION); to set a specific bit.

This changes only the chosen bit without affecting others.

Bit positions count from 0 at the rightmost bit.

Practice

1. What does the expression REGISTER |= (1 << 3); do in embedded C?

easy

A. Clears the 3rd bit (bit position 2) of REGISTER to 0

B. Sets the 4th bit (bit position 3) of REGISTER to 1 without changing other bits

C. Toggles all bits of REGISTER

D. Resets REGISTER to zero

Solution

1. Step 1: Understand bit shifting

   The expression 1 << 3 shifts the number 1 three places to the left, resulting in binary 00001000, which targets bit position 3.

2. Step 2: Apply bitwise OR to set bit

   Using |= with this mask sets bit 3 to 1 in REGISTER without changing other bits.

3. Final Answer:

   Sets the 4th bit (bit position 3) of REGISTER to 1 without changing other bits -> Option B

4. Quick Check:

   Bit shift + OR sets specific bit = A [OK]

Hint: Use |= with (1 << bit) to set that bit only [OK]

Common Mistakes:

• Confusing bit positions starting from 1 instead of 0
• Using & instead of | to set bits
• Thinking it clears bits instead of setting

2. Which of the following is the correct syntax to set bit 5 of a register named PORTA?

easy

A. PORTA = 1 << 5;

B. PORTA ^= (1 << 5);

C. PORTA &= ~(1 << 5);

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

Solution

1. Step 1: Identify correct operator to set bit

   To set a bit without affecting others, use bitwise OR assignment |= with a mask.

2. Step 2: Check the bit mask and syntax

   1 << 5 creates a mask for bit 5. Combining with |= sets that bit.

3. Final Answer:

   PORTA |= (1 << 5); -> Option D

4. Quick Check:

   Use |= with shifted 1 to set bit [OK]

Hint: Use |= with (1 << bit) to set bit correctly [OK]

Common Mistakes:

• Using = instead of |= overwrites all bits
• Using &= ~ to clear bits, not set
• Using ^= toggles bits instead of setting

3. Given unsigned char REG = 0x12; (binary 00010010), what is the value of REG after executing REG |= (1 << 4);?

medium

A. 0x32

B. 0x12

C. 0x16

D. 0x10

Solution

1. Step 1: Convert initial REG value to binary

   0x12 in binary is 00010010. Bit positions (0 to 7) from right to left.

2. Step 2: Calculate mask and apply OR

   1 << 4 is 00010000 (bit 4 set). OR with 00010010 results in 00110010 (0x32).

3. Final Answer:

   0x32 -> Option A

4. Quick Check:

   0x12 | 0x10 = 0x32 [OK]

Hint: Convert hex to binary, OR with mask, convert back [OK]

Common Mistakes:

• Miscounting bit positions starting from 1
• Confusing hex and decimal values
• Using &= or ^= instead of |=

4. Identify the error in the following code snippet intended to set bit 2 of REG:

unsigned char REG = 0x00;
REG = REG & (1 << 2);

medium

A. Using & instead of | clears all bits except bit 2

B. Bit shift operator << is used incorrectly

C. REG should be declared as int, not unsigned char

D. No error, code correctly sets bit 2

Solution

1. Step 1: Analyze bitwise AND usage

   The code uses & with mask (1 << 2), which clears all bits except bit 2.

2. Step 2: Understand effect on REG

   Since REG starts at 0, AND with mask results in 0, not setting bit 2. To set bit 2, OR |= should be used.

3. Final Answer:

   Using & instead of | clears all bits except bit 2 -> Option A

4. Quick Check:

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

Hint: Use |= to set bits, & clears bits [OK]

Common Mistakes:

• Confusing & and | operators
• Assuming &= sets bits
• Ignoring initial value of REG

5. You have a 16-bit register REG16 initially set to 0x0000. Write the correct statement to set bits 0, 7, and 15 simultaneously without affecting other bits.

hard

A. REG16 = (1 << 0) | (1 << 7) | (1 << 15);

B. REG16 = (1 << 0) & (1 << 7) & (1 << 15);

C. REG16 |= (1 << 0) | (1 << 7) | (1 << 15);

D. REG16 |= (1 << 0) & (1 << 7) & (1 << 15);

Solution

1. Step 1: Create mask for bits 0, 7, and 15

   Use bitwise OR to combine masks: (1 << 0) | (1 << 7) | (1 << 15) sets bits 0, 7, and 15.

2. Step 2: Use |= to set bits without clearing others

   Using |= applies the mask to REG16, setting those bits while preserving others.

3. Final Answer:

   REG16 |= (1 << 0) | (1 << 7) | (1 << 15); -> Option C

4. Quick Check:

   Combine masks with | and set with |= [OK]

Hint: Combine bit masks with | and set with |= [OK]

Common Mistakes:

• Using & instead of | to combine masks
• Using = overwrites all bits, losing previous data
• Using & instead of | in assignment