C Program to Swap Two Numbers Using XOR Operator

You can swap two numbers in C without a temporary variable using XOR operator like this: a = a ^ b; b = a ^ b; a = a ^ b; which swaps values of a and b.

Examples

Inputa=5, b=10

OutputAfter swap: a=10, b=5

Inputa=0, b=0

OutputAfter swap: a=0, b=0

Inputa=-3, b=7

OutputAfter swap: a=7, b=-3

How to Think About It

To swap two numbers using XOR, think of XOR as a way to combine bits without losing information. By applying XOR three times between the two numbers, you can exchange their values without needing extra space for a temporary variable.

Algorithm

1

Get input values for two numbers a and b

2

Apply XOR between a and b and store in a

3

Apply XOR between new a and b and store in b

4

Apply XOR between new a and new b and store in a

5

Print the swapped values of a and b

Code

c

#include <stdio.h>

int main() {
    int a = 5, b = 10;
    printf("Before swap: a=%d, b=%d\n", a, b);
    a = a ^ b;
    b = a ^ b;
    a = a ^ b;
    printf("After swap: a=%d, b=%d\n", a, b);
    return 0;
}

Output

Before swap: a=5, b=10
After swap: a=10, b=5

Dry Run

Let's trace swapping a=5 and b=10 through the XOR swap code

1

Initial values

a=5 (0101), b=10 (1010)

2

a = a ^ b

a = 0101 ^ 1010 = 1111 (15), b=10

3

b = a ^ b

b = 1111 ^ 1010 = 0101 (5), a=15

4

a = a ^ b

a = 1111 ^ 0101 = 1010 (10), b=5

Step	a (decimal)	b (decimal)
Initial	5	10
After a = a ^ b	15	10
After b = a ^ b	15	5
After a = a ^ b	10	5

Why This Works

Step 1: First XOR operation

When you do a = a ^ b, you store combined bits of both numbers in a without losing information.

Step 2: Second XOR operation

Then b = a ^ b extracts the original value of a and assigns it to b.

Step 3: Third XOR operation

Finally, a = a ^ b extracts the original value of b and assigns it to a, completing the swap.

Alternative Approaches

Using a temporary variable

c

#include <stdio.h>

int main() {
    int a = 5, b = 10, temp;
    printf("Before swap: a=%d, b=%d\n", a, b);
    temp = a;
    a = b;
    b = temp;
    printf("After swap: a=%d, b=%d\n", a, b);
    return 0;
}

This method is simple and clear but uses extra memory for a temporary variable.

Using addition and subtraction

c

#include <stdio.h>

int main() {
    int a = 5, b = 10;
    printf("Before swap: a=%d, b=%d\n", a, b);
    a = a + b;
    b = a - b;
    a = a - b;
    printf("After swap: a=%d, b=%d\n", a, b);
    return 0;
}

This method avoids extra variables but can cause overflow if numbers are large.

Complexity: O(1) time, O(1) space

Time Complexity

The swap uses a fixed number of operations (3 XORs), so it runs in constant time.

Space Complexity

No extra memory is used besides the input variables, so space complexity is constant.

Which Approach is Fastest?

All methods run in constant time; XOR swap uses no extra memory but is less readable than using a temporary variable.

Approach	Time	Space	Best For
XOR Swap	O(1)	O(1)	Memory-constrained environments
Temporary Variable	O(1)	O(1)	Readability and safety
Addition/Subtraction	O(1)	O(1)	Avoiding extra variables but watch overflow