Given a number x and two positions (from right side) in binary representation of x, write a function that swaps n bits at given two positions and returns the result. It is also given that the two sets of bits do not overlap.

Let p1 and p2 be the two given positions.

Example 1

Input:

x = 47 (00101111)

p1 = 1 (Start from second bit from right side)

p2 = 5 (Start from 6th bit from right side)

n = 3 (No of bits to be swapped)

Output:

227 (11100011)

The 3 bits starting from the second bit (from right side) are

swapped with 3 bits starting from 6th position (from right side)

Example 2

Input:

x = 28 (11100)

p1 = 0 (Start from first bit from right side)

p2 = 3 (Start from 4th bit from right side)

n = 2 (No of bits to be swapped)

Output:

7 (00111)

The 2 bits starting from 0th postion (from right side) are

swapped with 2 bits starting from 4th position (from right side)

**Solution**

We need to swap two sets of bits. XOR can be used in a similar way as it is used to swap 2 numbers. Following is the algorithm.

1) Move all bits of first set to rightmost side set1 = (x >> p1) & ((1U << n) - 1) Here the expression(1U << n) - 1gives a number that contains last n bits set and other bits as 0. We do & with this expression so that bits other than the last n bits become 0. 2) Move all bits of second set to rightmost side set2 = (x >> p2) & ((1U << n) - 1) 3) XOR the two sets of bits xor = (set1 ^ set2) 4) Put the xor bits back to their original positions. xor = (xor << p1) | (xor << p2) 5) Finally, XOR the xor with original number so that the two sets are swapped. result = x ^ xor

**Implementation:**

## C

// C program to illustrate..... #include<stdio.h> int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n) { /* Move all bits of first set to rightmost side */ unsigned int set1 = (x >> p1) & ((1U << n) - 1); /* Moce all bits of second set to rightmost side */ unsigned int set2 = (x >> p2) & ((1U << n) - 1); /* XOR the two sets */ unsigned int xor = (set1 ^ set2); /* Put the xor bits back to their original positions */ xor = (xor << p1) | (xor << p2); /* XOR the 'xor' with the original number so that the two sets are swapped */ unsigned int result = x ^ xor; return result; } /* Drier program to test above function*/ int main() { int res = swapBits(28, 0, 3, 2); printf("\nResult = %d ", res); return 0; }

## Java

// Java Program to illustrate... class GFG { static int swapBits(int x, int p1, int p2, int n) { // Move all bits of first set // to rightmost side int set1 = (x >> p1) & ((1 << n) - 1); // Moce all bits of second set //to rightmost side int set2 = (x >> p2) & ((1 << n) - 1); // XOR the two sets int xor = (set1 ^ set2); // Put the xor bits back to // their original positions xor = (xor << p1) | (xor << p2); // XOR the 'xor' with the original number // so that the two sets are swapped int result = x ^ xor; return result; } // Drier program public static void main(String[] args) { int res = swapBits(28, 0, 3, 2); System.out.println("Result = " + res); } } // This code is contributed by prerna saini.

Output:

Result = 7

Following is a shorter implementation of the same logic

int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n) { /* xor contains xor of two sets */ unsigned int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1); /* To swap two sets, we need to again XOR the xor with original sets */ return x ^ ((xor << p1) | (xor << p2)); }

**References:**

Swapping individual bits with XOR

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