Set or Clear a Specific Bit in a Register

Solving Approach

Clearing a bit needs BitwiseAND(&) operator so irrespective of existing bit in place, it clears when we operate with bitwise AND(&). Create a mask based on the bit position, say pos = 2, mask is ~(1<<2). Which means ~(0000 0100) = 1111 1011. So bitwise AND with this mask, can clear the bitposition 2 in given reg.

Likewise Setting a bit needs Bitwise OR(|) operator so it modifies the bit to 1 irrespective of the existing value inplace. Create a mask based on bit position, say pos = 2,mask is just (1<<2) i.e., 0000 0100.

Code

#include <stdio.h>

unsigned char modifyBit(unsigned char reg, int pos, int mode) {
    // Write your code here
    if(mode ==0){
       reg = reg & (~(1<<pos));

    }
    else if(mode == 1){
        reg = reg | (1<<pos) ;
    }
return reg;

}

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