How to rotate the bits of given number?

Question

How do I rotate the bits of given number, input number of rotations, at runtime?

For example:

binary:    10000000000000000000000000001011
rotations: 3 times right
result:    01110000000000000000000000000001

Similarly:

binary     10000000000000000000000000001011
rotations: 4 times left
result:    00000000000000000000000010111000

Below is my code to swap, but I couldn't find the logic to rotate the bits.

#include<stdio.h>
int main()
{
    int num,bitleft,bitright,i,j;
    printf("enter ur number\n");
    scanf("%d",&num);

    for(i=31,j=0;i>j;i--,j++) 
    {
        bitleft=num>>i&1;  //storing bits in integer from left
        bitright=num>>j&1;  //storing bits in integer from right
        if(bitleft!=bitright) // checking if bits are not similarly
        {
            num=num^1<<i; // not similar then complement
            num=num^1<<j;

        }
    }
    printf("\n");

    for(j=31;j>=0;j--)  // loop to print swapped bits
    {
        if(num&1<<j)
            printf("1");
        else
            printf("0");
    } 
}

@user3859409 What does this words "no of rotations ,at runtime" mean? — Vlad from Moscow, Aug 11 '14 at 18:27
[Wikipedia](http://en.wikipedia.org/wiki/Circular_shift#Implementing_circular_shifts) provides a nice clean implementation — Drew McGowen, Aug 11 '14 at 18:27
DR; Maybe related: http://stackoverflow.com/questions/776508/circular-shift-operations-in-c — 101010, Aug 11 '14 at 18:28
To rotate `num` to the left by `k` bits, write `(num << k) | (num >> (sizeof(num)*CHAR_BIT - k))`. — leemes, Aug 11 '14 at 18:32
@fred Larson i know it is asked before but answer are with built in function i don't want to use any built in function and this was the qustion asked in google interview ? — user3859409, Aug 11 '14 at 18:47
@user3859409: There are 13 answers there. Some use inline asm, some use intrinsic functions, some are plain code snippets. What more are you looking for? — Fred Larson, Aug 11 '14 at 18:52

NetVipeC · Answer 1 · 2014-08-11T18:56:05.450

Try this:

#include <iostream>

template <typename T>
T rol_(T value, int count) {
    return (value << count) | (value >> (sizeof(T)*CHAR_BIT - count));
}

template <typename T>
T ror_(T value, int count) {
    return (value >> count) | (value << (sizeof(T)*CHAR_BIT - count));
}

int main() {
    unsigned int a = 0x8000000B; // 10000000000000000000000000001011 in binary
    std::cout << "A = 0x" << std::hex << a << ", ror(A, 3) = 0x" << ror_(a, 3) << std::endl;
    std::cout << "A = 0x" << std::hex << a << ", rol(A, 3) = 0x" << rol_(a, 3) << std::endl;
    return 0;
}

Notes:

(value << count) in rol and (value >> count) in ror set the high bits in rol and the lower bits in ror. The other bits fall away.
(value >> (sizeof(T)*CHAR_BIT - count)) in rol and (value << (sizeof(T)*CHAR_BIT - count)) in ror set the bits that fall away in the previous operation.

Example (assuming types of 32bits):

binary:               10000000000000000000000000001011
rotations: 3 times right
(value >> 3):         00010000000000000000000000000001
(value << (32 - 3)):  01100000000000000000000000000000
------------------------------------------------------
                      01110000000000000000000000000001

i was looking for solution without using any inbuilt function or any library — user3859409, Aug 11 '14 at 18:51

How to rotate the bits of given number?

1 Answers1