Set Multiple Bits in 8-bit Register

Code

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

uint8_t set_range(uint8_t reg, uint8_t start, uint8_t end) {
    // Your code here
    uint8_t len = end - start + 1;
    uint8_t mask =((1<<len)-1)<<start;
    return reg | mask;
}

int main() {
    uint8_t reg, start, end;
    scanf("%hhu %hhu %hhu", &reg, &start, &end);
    printf("%u", set_range(reg, start, end));
    return 0;
}

Solving Approach

my approach here is , first we find the length of bits needed to make 1

we do by the following 

• length = end - start +1; (if end =3 , start =0 , len = 4(as per 0 indexing)

• to find the mask ,

   - 2^4 = 16 , 16-1=15=0x1111, so this gives us the mask of 4 bits,

  - rather than using the powe function we take help of shift, as 

  • 1<<4 = 2**4 = 2^4 = 16

  • then obtained mask is shifted by start bit 

  • mask is or with the register.