How to swap two numbers without using a temporary variable?

Given two variables, x and y, swap two variables without using a third variable.

Method 1 (Using Arithmetic Operators)
The idea is to get sum in one of the two given numbers. The numbers can then be swapped using the sum and subtraction from sum.

#include <stdio.h>

int main()

{

  int x = 10, y = 5;

  // Code to swap 'x' and 'y'

  x = x + y;  // x now becomes 15

  y = x - y;  // y becomes 10

  x = x - y;  // x becomes 5

  printf("After Swapping: x = %d, y = %d", x, y);

  return 0;

output:

After Swapping: x = 5, y = 10

Multiplication and division can also be used for swapping.

#include <stdio.h>

int main()

{

  int x = 10, y = 5;

  // Code to swap 'x' and 'y'

  x = x * y;  // x now becomes 50

  y = x / y;  // y becomes 10

  x = x / y;  // x becomes 5

  printf("After Swapping: x = %d, y = %d", x, y);

  return 0;

output:

After Swapping: x = 5, y = 10

Method 2 (Using Bitwise XOR)
The bitwise XOR operator can be used to swap two variables. The XOR of two numbers x and y returns a number which has all the bits as 1 wherever bits of x and y differ. For example, XOR of 10 (In Binary 1010) and 5 (In Binary 0101) is 1111 and XOR of 7 (0111) and 5 (0101) is (0010).

#include <stdio.h>

int main()

{

  int x = 10, y = 5;

  // Code to swap 'x' (1010) and 'y' (0101)

  x = x ^ y;  // x now becomes 15 (1111)

  y = x ^ y;  // y becomes 10 (1010)

  x = x ^ y;  // x becomes 5 (0101)

  printf("After Swapping: x = %d, y = %d", x, y);

  return 0;Output:

After Swapping: x = 5, y = 10

Problems with above methods
1) The multiplication and division based approach doesn’ work if one of the numbers is 0 as the product becomes 0 irrespective of the other number.

2) Both Arithmetic solutions may cause arithmetic overflow. If x and y are too large, addition and multiplication may go out of integer range.

3) When we use pointers to variable and make a function swap, all of the above methods fail when both pointers point to the same variable. Let’s take a look what will happen in this case if both are pointing to the same variable.

// Bitwise XOR based method
x = x ^ x; // x becomes 0
x = x ^ x; // x remains 0
x = x ^ x; // x remains 0

// Arithmetic based method
x = x + x; // x becomes 2x
x = x – x; // x becomes 0
x = x – x; // x remains 0

Let us see the following program.

#include <stdio.h>

void swap(int *xp, int *yp)

{

    *xp = *xp ^ *yp;

    *yp = *xp ^ *yp;

    *xp = *xp ^ *yp;

}

int main()

{

  int x = 10;

  swap(&x, &x);

  printf("After swap(&x, &x): x = %d", x);

  return 0;

Output:

After swap(&x, &x): x = 0

Swapping a variable with itself may need in many standard algorithms. For example, see this implementation of QuickSort where we may swap a variable with itself. The above problem can be avoided by putting a condition before the swapping.

#include <stdio.h>

void swap(int *xp, int *yp)

{

    if (xp == yp) // Check if the two addresses are same

      return;

    *xp = *xp + *yp;

    *yp = *xp - *yp;

    *xp = *xp - *yp;

}

int main()

{

  int x = 10;

  swap(&x, &x);

  printf("After swap(&x, &x): x = %d", x);

  return 0;

Output: