Rotate Right in a 32-bit Register

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

uint32_t rotate_right(uint32_t reg, uint8_t n) {
    // Your code here
    reg=(reg>>n)|(reg<<32-n);
    return reg;
}

int main() {
    uint32_t reg;
    uint8_t n;
    scanf("%u %hhu", &reg, &n);
    printf("%u", rotate_right(reg, n));
    return 0;
}

the given register is a 32 bit in size
step1:reg>>n(no ,. of times the register has to be shifted)

step2:(extract the last 4 bits to position msb's)

step3: bitwise or(|) answers of step 1 and step 2
 

the answer is right shifted by n times circlar buffer