Rotate Left in an 8-bit Register

Code

#include <stdio.h>
#include <stdint.h>

#define himask 0x80
#define lomask 0x01

uint8_t rotate_left(uint8_t reg, uint8_t n) {
    // Your code here
    uint8_t i;
    for (i = 0; i < n; i++)
    {
        if (reg & himask) {
            reg = ( reg << 1 ) | 1;
        }
        else {
            reg = reg << 1;
        }
    }
    return reg;
}

int main() {
    uint8_t reg, n;
    scanf("%hhu %hhu", &reg, &n);
    printf("%u", rotate_left(reg, n));
    return 0;
}

Solving Approach

use a mask to test the msb before you shift it left.

If the test is zero then just shift it left

else shift it left and introduce a 1 in lsb