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 needed 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:

After swap(&x, &x): x = 10

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