Bit rotations exercise

Question

I'm studying C on K&R and I solved exercise 2.08:

Write a function rightrot(x,n) that returns the value of the integer x rotated to the right by n positions

I've tested my code with some bit patterns and it seems to work, but I'm not sure that this solution covers all possible cases.

What do you think about this code?

unsigned rightrot(unsigned x, int n)
{   
    int size;
    unsigned y;

    size = 0;
    y = x;

    while (y != 0) {
    y = y << BYTESIZE;
    size++;
    }
    size = size * BYTESIZE;

    return (x << (size-n)) | (x >> n);
}

Here is the main

#include <stdio.h>

#define BYTESIZE 8

unsigned rightrot(unsigned x, int n);

int main(void)
{
     unsigned x;
     int n;

     x = 0x23acb;
     n = 2;

     printf("%x\n", rightrot(x, n));

     return 0;
}

Answer 1

@William has the optimum solution.

But some comments on your code:

// There is already a macro that defines the number of bits
// in a byte it is called CHAR_BIT
#define BYTESIZE 8

    // Initialize variables as you declare them   
    int size;
    unsigned y;

    // You can find the number of bytes in an object using
    // sizeof(<expression>)    
    while (y != 0) {
    y = y << BYTESIZE;
    size++;
    }

    // Thus the number of bits is:
    // sizeof(y) * CHAR_BIT
    size = size * BYTESIZE;

    // The result is the same.
    return (x << (size-n)) | (x >> n);

Answer 2

Isn't it something like this?

#include <limits.h> /* for CHAR_BIT */

unsigned
rotate_right(unsigned x, int n)
{
    int left_shift = ((sizeof x) * CHAR_BIT) - n;
    return x<<left_shift | x>>n;
}