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;
}

• cimere: I'm aware of the existence of built in sizeof function but I can not use it because formally in the book it has not yet been explained. :-) Is it correct to calculate the size of x as I did or am I missing something?
• William Morris: No. In your solution, if x==0, size is 0. Try for(size=0, y=~0; y; y<<=1, size++); (note the ; at the end of that)
• edmz: +1. GCC 4.9 can optimize it to a simple ror. Check here. Lower versions can't though, even with higher optimization levels.