C Bitwise Circle Shifting - Unexpected Results - c

Question

I'm currently working through the K&R C book exercises and am on exercise 8 of chapter 2. The challenge is to write a function 'rotright' that rotates (or circle shifts) the bits of unsigned integer x by n bits. I believe I've come up with a solution, but it's not returning what I would expect. Given the number 213, which is 11010101 in binary, rotating 2 bits to the right would yield 01110101, which is 117. However, my program upon being given x=213 and n=2 returns 53. I've tried writing out the process of what's happening to the integer in binary form in comments and can't find a problem. Any help would be appreciated.
#include <stdio.h>
unsigned rotright(unsigned x, int n)
{
/* Example with x = 11010101 (213 in decimal), n = 2
First iteration:
x = (01101010) | ~(11111111 >> 1) = 11101010
Second iteration:
x = (01110101) | ~(11111111 >> 0) = 01110101
Returns 01110101
right shifts only if last bit of x == 1, then sets first bit of right shifted x to 1
if last bit of x == 0, x is right shifted by 1 and then unchanged.
(01101010) | ~(11111111 >> (11010101 & 00000001))
= 01101010 | ~(11111111 >> 00000001)
= 01101010 | 10000000 = 11101010
(11101010) | ~(11111111 >> (11101010 & 00000001))
= 01110101 | ~(11111111 >> 0)
= 01110101 | 00000000 = 01110101
*/
for (; n > 0; n--)
x = (x >> 1) | ~(~0 >> (x & 1));
return x;
}
int main()
{
printf("%d\n", rotright(213, 2));
return 0;
}

Answer 1

x = (x >> 1) | ~(~0 >> (x & 1));
you get 53 because this is (213 >> 2)
~(~0 >> (x & 1)) is always 0, because ~0 is -1, and (-1 >> n) is again -1 in your case, and finally ~(-1) is 0
You want that :
#include <stdio.h>
#include <limits.h>
unsigned rotright(unsigned x, int n)
{
unsigned mask = (1u << (CHAR_BIT * sizeof(int) - 1));
for (; n > 0; n--) {
x = (x / 2) | ((x & 1) ? mask : 0);
}
return x;
}
int main()
{
printf("%d\n", rotright(213, 2));
return 0;
}
On 32bits the result is 1073741877 being 1000000000000000000000000110101, not 117 whose supposes you work on 8 bits