Exercise 2.8-returns the value of the integer x rotated to the right by n positions

Question

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

/* write a function rightrot(x,n) that returns the value of the integer x rotated to rightby n bit positions */

#include<stdio.h>

unsigned rightrot(unsigned x,int n);

int main(void)
{
    printf("%u",(unsigned)rightrot((unsigned)8,(int)1));

    return 0;
}

/* rightrot: rotate x to right by n bit positions */

unsigned rightrot(unsigned x,int n)
{
    int wordlength(void);
    unsigned rbit;/* rightmost bit */
    
    rbit = x << (wordlength() - n);
    x = x >> n;
    x = x | rbit;
    
    return x;
}

int wordlength(void)
{
    int i;
    unsigned v = (unsigned) ~0;

    for(i=1;(v=v>>1)>0;i++)
        ;
    return i;
}

        
Run this

Explanation

We need to get the right most bit of the number provided.

First we get the right n bits at one time.

::
rbit = x << (wordlength() - n);

Once we get the right n bits, in order to rotate the value of x, we right shift x for n bits and then OR the result of x with the rbit determined in the previous step.

       x = x >> n;
       x = x | rbit;

For the same example.
n is between 0 - wordlength()

condition 1.when  (n == 0) or (n == wordlength())

            rightrot(x, 0) == x

condition 2. when  (n > 0) and (n < wordlength())   like n = 3

            x  = 0001 1001
            the right n bits will be 001.
            the right rightrot(x,n)result should be 0010 0011

            x << (wordlength() - n) = 0001 1001 << (8 - 3)
            = 0001 1001 << 5
            = 0010 0000

So we have got the right most n bits set.Now we right x by 1 and OR the rbit with x.

x >> n =  0001 1001 >> 3
=  0000 0011

x | rbit = 0000 0011 | 0010 0000
= 0010 0011

Which is our expected result.

condition 3. when (n > wordlength())        like n = 12

The Compiler will auto transfer “n” to “n % wordlength()”, n will be 3, then see “n” as condition 2. The result should be correct too!

::
condition 4. when n < 0 (which is not Often use)

The result will mirror the function,the rightrot(x,n) function will move the left most n(n > 0)bits to the right side ,the function should called leftrot(x,n).

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