Rotate bits of a number

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    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<>(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 << " Right Rotation of " << n <<             " by " << d << " is ";     cout << g.rightRotate(n, d);     getchar(); }     // This code is contributed by SoM15242

C

 #include #define INT_BITS 32    /*Function to left rotate n by d bits*/ int leftRotate(int n, unsigned int d) {    /* In 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(" Right Rotation of %d by %d is ", n, d);   printf("%d", rightRotate(n, d));   getchar(); }

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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<>(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(" Right 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<>(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<>(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(" Right Rotation of " + n                         + " by " + d + " is ");         Console.Write(rightRotate(n, d));     } }    // This code is contributed by Sam007

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.