Set or Clear a Specific Bit in a Register

Code

#include <stdio.h>

unsigned char modifyBit(unsigned char reg, int pos, int mode) {
    if (mode == 1) {
        // Set the bit at position 'pos'
        return reg | (1 << pos);
    } else {
        // Clear the bit at position 'pos'
        return reg & ~(1 << pos);
    }
}



int main() {
    unsigned char reg;
    int pos, mode;
    scanf("%hhu %d %d", &reg, &pos, &mode);
    printf("%d", modifyBit(reg, pos, mode));
    return 0;
}

Solving Approach

An 8-bit register holds values from 0 to 255, represented in binary as 00000000 to 11111111. Each bit position (0 to 7) can be individually manipulated using bitwise operations.

🛠️ Step-by-Step Solving Approach

1. Input Handling
   • Read three values from the user:
     • reg: the 8-bit register value (unsigned char)
     • pos: the bit position to modify (0–7)
     • mode: operation mode (1 to set, 0 to clear)
2. Validation (Optional but recommended)
   • Ensure pos is between 0 and 7
   • Ensure mode is either 0 or 1
3. Bit Manipulation Logic
   • Use bitwise operators to modify the bit:
     • Set a bit: reg | (1 << pos)
       • Shifts 1 left by pos bits to create a mask
       • ORs the mask with reg to set the bit
     • Clear a bit: reg & ~(1 << pos)
       • Shifts 1 left by pos bits to create a mask
       • Inverts the mask with ~ to get all 1s except the target bit
       • ANDs the inverted mask with reg to clear the bit
4. Return or Print the Result
   • Output the modified register value in decimal

🔍 Example Walkthrough

Input: reg = 10, pos = 3, mode = 1

• Binary of 10: 00001010
• 1 << 3 → 00001000
• reg | 00001000 → 00001010 → Decimal 10 (bit already set)

Input: reg = 10, pos = 1, mode = 0

• Binary of 10: 00001010
• 1 << 1 → 00000010
• ~00000010 → 11111101
• reg & 11111101 → 00001000 → Decimal 8