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;
}
```

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).