Set or Clear a Specific Bit in a Register

Solving Approach

First I select the mode, do I need to set bit or clear bit.

Setting bit is easy-er. I start with the (1 << pos) part. This part of the code means that I take number 1 and shift it to the left the same amount of times as the variable pos. If i convert the number 1 into binery it would look like this 00000001. I can imagine that the pos = 3, (00000001 << 3)so the number 1 will be shifted 3 times to the left giving us the result of 00001000. Perfect...now the juicy part!

The number in reg for example 10 in binary is 00001010

reg |=  (1 << pos)    ==     10|=  (1 << 3)      ==    00001010|=  ( 00000001 <<  3) 

This command |= will do the last lift. It will do the math and store the value in its self. So it could be written as well like this " reg = reg |  (1 << pos) ". Then its only the matter of fact .....what dose the | thingy dose? Well, it is a OR operand, in short:

0001 <- first number            0000   0001   0000

0001 <- second number      0001   0000   0000

====                                      =====   ====   ====

0001 <-result                        0001    0001   0000

If one of the numbers are 1 then the result will be 1.

In our case:

00001010

00001000

========

00001010

Result didnt cahnge becouse there was already 1.

Something similar is happening in mode 0, but it has the opposite meaning. If it finds a matching 1, it will write 0 in its place.

00001010

00001000

========

00000010

Hope it helped out someone :D

Code

#include <stdio.h>

unsigned char modifyBit(unsigned char reg, int pos, int mode) {
    if(mode == 1){
        reg |=  (1 << pos);
        return reg;
    }else if(mode == 0){
        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;
}