Swap using XOR

Swap Two Numbers using XOR in C++

IntermediateTopic: Bitwise Operations Programs
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C++ Swap using XOR Program

This program helps you to learn the fundamental structure and syntax of C++ programming.

Try This Code
#include <iostream>
using namespace std;

void swapXOR(int& a, int& b) {
    a = a ^ b;
    b = a ^ b;  // b = (a^b) ^ b = a
    a = a ^ b;  // a = (a^b) ^ a = b
}

int main() {
    int a = 10, b = 20;
    
    cout << "Before swap: a = " << a << ", b = " << b << endl;
    
    swapXOR(a, b);
    
    cout << "After swap: a = " << a << ", b = " << b << endl;
    
    // Multiple swaps
    cout << "\nMultiple swaps:" << endl;
    int x = 5, y = 15, z = 25;
    
    cout << "Before: x = " << x << ", y = " << y << ", z = " << z << endl;
    
    swapXOR(x, y);
    swapXOR(y, z);
    
    cout << "After: x = " << x << ", y = " << y << ", z = " << z << endl;
    
    // XOR properties demonstration
    cout << "\nXOR properties:" << endl;
    int num1 = 12, num2 = 25;
    cout << "num1 = " << num1 << ", num2 = " << num2 << endl;
    cout << "num1 ^ num2 = " << (num1 ^ num2) << endl;
    cout << "(num1 ^ num2) ^ num2 = " << ((num1 ^ num2) ^ num2) << " (back to num1)" << endl;
    cout << "(num1 ^ num2) ^ num1 = " << ((num1 ^ num2) ^ num1) << " (back to num2)" << endl;
    
    return 0;
}
Output
Before swap: a = 10, b = 20
After swap: a = 20, b = 10

Multiple swaps:
Before: x = 5, y = 15, z = 25
After: x = 15, y = 25, z = 5

XOR properties:
num1 = 12, num2 = 25
num1 ^ num2 = 21
(num1 ^ num2) ^ num2 = 12 (back to num1)
(num1 ^ num2) ^ num1 = 25 (back to num2)

Understanding Swap using XOR

This program teaches you how to Swap Two Numbers using XOR in C++. The XOR swap algorithm is an elegant bit manipulation technique that swaps two values without using a temporary variable. It demonstrates the unique properties of the XOR operation.

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1. What This Program Does

The program demonstrates XOR swap algorithm:

•Swapping two numbers without temporary variable
•Using XOR properties
•Multiple swaps demonstration
•XOR properties explanation

XOR swap is a clever bit manipulation technique.

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2. Header Files Used

1.#include <iostream>
•Provides cout and cin for input/output operations.

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3. Understanding XOR Swap

XOR Properties

:

•a ^ a = 0 (XOR with itself is zero)
•a ^ 0 = a (XOR with zero is identity)
•(a ^ b) ^ b = a (XOR is reversible)
•Commutative: a ^ b = b ^ a

Swap Algorithm

:

•Uses XOR properties
•No temporary variable needed
•Three XOR operations
•Elegant solution

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4. XOR Swap Algorithm

Three Steps

:

1.a = a ^ b
2.b = a ^ b // b = (a^b) ^ b = a
3.a = a ^ b // a = (a^b) ^ a = b

How it works

:

•Step 1: Store XOR in a
•Step 2: Recover original a in b
•Step 3: Recover original b in a
•Swaps values using XOR properties

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5. Why It Works

Mathematical Proof

:

•After step 1: a = a^b, b = b
•After step 2: a = a^b, b = (a^b)^b = a
•After step 3: a = (a^b)^a = b, b = a
•Result: a and b are swapped

XOR Reversibility

:

•(a ^ b) ^ b = a (recover a)
•(a ^ b) ^ a = b (recover b)
•XOR is its own inverse

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6. Important Limitation

Same Memory Location

:

•Don't use if variables are same
•a ^ a = 0 (would zero the value)
•Always check if different
•Use temporary for safety

Example Problem

:

•swapXOR(x, x) would set x to 0
•Must ensure different variables
•Check before swapping

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7. When to Use XOR Swap

Best For

:

•Embedded systems (limited memory)
•Educational purposes
•Demonstrating XOR properties
•Memory-constrained environments

Not Recommended For

:

•Modern systems (compiler optimizes)
•Readability concerns
•Same variable risk
•Production code (usually)

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8. Modern Compilers

Compiler Optimization

:

•Modern compilers optimize swaps
•Temporary variable optimized away
•XOR swap may be slower
•Use standard swap for clarity

Recommendation

:

•Use std::swap() in production
•XOR swap for learning
•Understand the concept
•Prefer readability

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9. XOR Properties Demonstration

Reversibility

:

•(a ^ b) ^ b = a
•(a ^ b) ^ a = b
•XOR can recover original values
•Foundation of XOR swap

Understanding

:

•Helps understand algorithm
•Demonstrates XOR power
•Educational value
•Foundation for other algorithms

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10. return 0;

This ends the program successfully.

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Summary

•XOR swap: swaps two numbers without temporary variable using XOR properties.
•Algorithm: a = a^b, b = a^b (recover a), a = a^b (recover b).
•Works because: (a^b)^b = a and (a^b)^a = b (XOR reversibility).
•Important: don't use if variables point to same memory location.
•Understanding XOR swap demonstrates XOR properties and bit manipulation.
•Useful in embedded systems, though modern compilers optimize standard swaps.

This program is fundamental for learning XOR properties, understanding bit manipulation techniques, and preparing for advanced bit manipulation algorithms in C++ programs.

Let us now understand every line and the components of the above program.

Note: To write and run C++ programs, you need to set up the local environment on your computer. Refer to the complete article Setting up C++ Development Environment. If you do not want to set up the local environment on your computer, you can also use online IDE to write and run your C++ programs.

Practical Learning Notes for Swap using XOR

This C++ program is part of the "Bitwise Operations Programs" topic and is designed to help you build real problem-solving confidence, not just memorize syntax. Start by understanding the goal of the program in plain language, then trace the logic line by line with a custom input of your own. Once you can predict the output before running the code, your understanding becomes much stronger.

A reliable practice pattern is to run the original version first, then modify only one condition or variable at a time. Observe how that single change affects control flow and output. This deliberate style helps you understand loops, conditions, and data movement much faster than copying full solutions repeatedly.

For interview preparation, explain this solution in three layers: the high-level approach, the step-by-step execution, and the time-space tradeoff. If you can teach these three layers clearly, you are ready to solve close variations of this problem under time pressure.

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