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Rotate bits of a number
  • Difficulty Level : Easy
  • Last Updated : 23 Apr, 2021

Bit Rotation: A rotation (or circular shift) is an operation similar to shift except that the bits that fall off at one end are put back to the other end. 
In left rotation, the bits that fall off at left end are put back at right end. 
In right rotation, the bits that fall off at right end are put back at left end.
 

Example: 
Let n is stored using 8 bits. Left rotation of n = 11100101 by 3 makes n = 00101111 (Left shifted by 3 and first 3 bits are put back in last ). If n is stored using 16 bits or 32 bits then left rotation of n (000…11100101) becomes 00..0011100101000. 
Right rotation of n = 11100101 by 3 makes n = 10111100 (Right shifted by 3 and last 3 bits are put back in first ) if n is stored using 8 bits. If n is stored using 16 bits or 32 bits then right rotation of n (000…11100101) by 3 becomes 101000..0011100. 
 

C++




// C++ code to rotate bits
// of number
#include<iostream>
 
using namespace std;
#define INT_BITS 32
class gfg
{
     
/*Function to left rotate n by d bits*/
public:
int leftRotate(int n, unsigned int d)
{
     
    /* In n<<d, last d bits are 0. To
     put first 3 bits of n at
    last, do bitwise or of n<<d
    with n >>(INT_BITS - d) */
    return (n << d)|(n >> (INT_BITS - d));
}
 
/*Function to right rotate n by d bits*/
int rightRotate(int n, unsigned int d)
{
    /* In n>>d, first d bits are 0.
    To put last 3 bits of at
    first, do bitwise or of n>>d
    with n <<(INT_BITS - d) */
    return (n >> d)|(n << (INT_BITS - d));
}
};
 
/* Driver code*/
int main()
{
    gfg g;
    int n = 16;
    int d = 2;
    cout << "Left Rotation of " << n <<
            " by " << d << " is ";
    cout << g.leftRotate(n, d);
    cout << "\nRight Rotation of " << n <<
            " by " << d << " is ";
    cout << g.rightRotate(n, d);
    getchar();
}
 
// This code is contributed by SoM15242
C



#include<stdio.h>
#define INT_BITS 32
 
/*Function to left rotate n by d bits*/
int leftRotate(int n, unsigned int d)
{
   /* In n<<d, last d bits are 0. To put first 3 bits of n at
     last, do bitwise or of n<<d with n >>(INT_BITS - d) */
   return (n << d)|(n >> (INT_BITS - d));
}
 
/*Function to right rotate n by d bits*/
int rightRotate(int n, unsigned int d)
{
   /* In n>>d, first d bits are 0. To put last 3 bits of at
     first, do bitwise or of n>>d with n <<(INT_BITS - d) */
   return (n >> d)|(n << (INT_BITS - d));
}
 
/* Driver program to test above functions */
int main()
{
  int n = 16;
  int d = 2;
  printf("Left Rotation of %d by %d is ", n, d);
  printf("%d", leftRotate(n, d));
  printf("\nRight Rotation of %d by %d is ", n, d);
  printf("%d", rightRotate(n, d));
  getchar();
}
Java



// Java code to rotate bits
// of number
class GFG
{
static final int INT_BITS = 32;
 
/*Function to left rotate n by d bits*/
static int leftRotate(int n, int d) {
     
    /* In n<<d, last d bits are 0.
       To put first 3 bits of n at
       last, do bitwise or of n<<d with
       n >>(INT_BITS - d) */
    return (n << d) | (n >> (INT_BITS - d));
}
 
/*Function to right rotate n by d bits*/
static int rightRotate(int n, int d) {
     
    /* In n>>d, first d bits are 0.
       To put last 3 bits of at
       first, do bitwise or of n>>d
       with n <<(INT_BITS - d) */
    return (n >> d) | (n << (INT_BITS - d));
}
 
// Driver code
public static void main(String arg[])
{
    int n = 16;
    int d = 2;
    System.out.print("Left Rotation of " + n +
                          " by " + d + " is ");
    System.out.print(leftRotate(n, d));
     
    System.out.print("\nRight Rotation of " + n +
                             " by " + d + " is ");
    System.out.print(rightRotate(n, d));
}
}
 
// This code is contributed by Anant Agarwal.
Python3



# Python3 code to
# rotate bits of number
 
INT_BITS = 32
 
# Function to left
# rotate n by d bits
def leftRotate(n, d):
 
    # In n<<d, last d bits are 0.
    # To put first 3 bits of n at
    # last, do bitwise or of n<<d
    # with n >>(INT_BITS - d)
    return (n << d)|(n >> (INT_BITS - d))
 
# Function to right
# rotate n by d bits
def rightRotate(n, d):
 
    # In n>>d, first d bits are 0.
    # To put last 3 bits of at
    # first, do bitwise or of n>>d
    # with n <<(INT_BITS - d)
    return (n >> d)|(n << (INT_BITS - d)) & 0xFFFFFFFF
 
# Driver program to
# test above functions
n = 16
d = 2
 
print("Left Rotation of",n,"by"
      ,d,"is",end=" ")
print(leftRotate(n, d))
 
print("Right Rotation of",n,"by"
     ,d,"is",end=" ")
print(rightRotate(n, d))
 
# This code is contributed by
# Smitha Dinesh Semwal
C#



// C# program to rotate
// bits of a number
using System;
 
class GFG
{
    static int INT_BITS = 32;
 
    /* Function to left rotate n by d bits*/
    static int leftRotate(int n, int d) {
         
        /* In n<<d, last d bits are 0.
        To put first 3 bits of n at
        last, do bitwise or of n<<d with
        n >>(INT_BITS - d) */
        return (n << d) | (n >> (INT_BITS - d));
    }
     
    /*Function to right rotate n by d bits*/
    static int rightRotate(int n, int d) {
         
        /* In n>>d, first d bits are 0.
        To put last 3 bits of at
        first, do bitwise or of n>>d
        with n <<(INT_BITS - d) */
        return (n >> d) | (n << (INT_BITS - d));
    }
     
    // Driver code
    public static void Main()
    {
        int n = 16;
        int d = 2;
         
        Console.Write("Left Rotation of " + n
                      + " by " + d + " is ");
        Console.Write(leftRotate(n, d));
         
        Console.Write("\nRight Rotation of " + n
                       + " by " + d + " is ");
        Console.Write(rightRotate(n, d));
    }
}
 
// This code is contributed by Sam007
Javascript



<script>
 
// JavaScript code to rotate bits
// of number
let INT_BITS = 32;
 
     
/*Function to left rotate n by d bits*/
 
function leftRotate( n,  d)
{   
    /* In n<<d, last d bits are 0. To
     put first 3 bits of n at
    last, do bitwise or of n<<d
    with n >>(INT_BITS - d) */
    return (n << d)|(n >> (INT_BITS - d));
}
 
/*Function to right rotate n by d bits*/
function rightRotate( n, d)
{
    /* In n>>d, first d bits are 0.
    To put last 3 bits of at
    first, do bitwise or of n>>d
    with n <<(INT_BITS - d) */
    return (n >> d)|(n << (INT_BITS - d));
}
 
 
/* Driver code*/
let n = 16;
let d = 2;
document.write("Left Rotation of " + n +
            " by " + d + " is ");
document.write(leftRotate(n, d));
document.write("<br>");           
document.write("\nRight Rotation of " + n +
            " by " + d + " is ");
document.write(rightRotate(n, d));
 
</script>
Output : 
Left Rotation of 16 by 2 is 64
Right Rotation of 16 by 2 is 4
 


Please write comments if you find any bug in the above program or other ways to solve the same problem. 
 
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