Add Two Numbers Program

Adding two numbers is one of the most basic operations in C++. It is commonly used to demonstrate arithmetic operators and different approaches to performing addition.

• Uses arithmetic and bitwise operations to compute the sum of two integers.
• Demonstrates multiple techniques for adding numbers in C++.
• Suitable for understanding basic programming logic.

Example: The simplest way to add two numbers is by using the addition operator (+).

#include <iostream>
using namespace std;

int main() {

    int a = 11, b = 9;

    cout << a + b;

    return 0;
}

#include <stdio.h>

int main() {
    int a = 11, b = 9;

    // Adding the two numbers and printing there sum
    printf("%d", a + b);

    return 0;
}

public class Main {
    public static void main(String[] args) {
        int a = 11, b = 9;

        // Adding the two numbers and printing there sum
        System.out.println(a + b);
    }
}

a = 11
b = 9

# Adding the two numbers and printing there sum
print(a + b)

using System;

class Program {
    static void Main() {
        int a = 11, b = 9;

        // Adding the two numbers and printing there sum
        Console.WriteLine(a + b);
    }
}

let a = 11, b = 9;

// Adding the two numbers and printing there sum
console.log(a + b);

20

Explanation: The + operator adds the values of a and b and displays the result.

Different Methods to Add Two Numbers

Add Two Numbers Using the Increment Operator (++)

Two numbers can also be added by repeatedly incrementing or decrementing one number based on the value of the other.

#include <iostream>
using namespace std;

int main() {

    int a = 11, b = 9;

    for (int i = 0; i < b; i++)
        a++;

    for (int i = 0; i > b; i--)
        a--;

    cout << a;

    return 0;
}

#include <stdio.h>

int main() {
    int a = 11, b = 9;

    // If b is positive, increment a to b times
    for (int i = 0; i < b; i++)
        a++;

    // If b is negative, decrement a to |b| times
    for (int i = 0; i > b; i--)
        a--;

    printf("%d", a);

    return 0;
}

public class Main {
    public static void main(String[] args) {
        int a = 11, b = 9;

        // If b is positive, increment a to b times
        for (int i = 0; i < b; i++)
            a++;

        // If b is negative, decrement a to |b| times
        for (int i = 0; i > b; i--)
            a--;

        System.out.println(a);
    }
}

def main():
    a = 11
    b = 9

    # If b is positive, increment a to b times
    for i in range(b):
        a += 1

    # If b is negative, decrement a to |b| times
    for i in range(-b):
        a -= 1

    print(a)

if __name__ == "__main__":
    main()

using System;

class Program {
    static void Main() {
        int a = 11, b = 9;

        // If b is positive, increment a to b times
        for (int i = 0; i < b; i++)
            a++;

        // If b is negative, decrement a to |b| times
        for (int i = 0; i > b; i--)
            a--;

        Console.WriteLine(a);
    }
}

function main() {
    let a = 11, b = 9;

    // If b is positive, increment a to b times
    for (let i = 0; i < b; i++)
        a++;

    // If b is negative, decrement a to |b| times
    for (let i = 0; i > b; i--)
        a--;

    console.log(a);
}

main();

20

Explanation: If b is positive, a is incremented b times. If b is negative, a is decremented |b| times.

Add Two Numbers Using Bitwise Operators

Addition can also be performed using bitwise operations without using the + operator.

Working

• XOR (^) computes the sum of bits without considering the carry.
• AND (&) computes the carry bits.
• The carry is left shifted by one position.
• The process is repeated until no carry remains.

#include <iostream>
using namespace std;

int main() {

    int a = 11, b = 9;

    while (b != 0) {

        int carry = a & b;

        a = a ^ b;

        b = carry << 1;
    }

    cout << a;

    return 0;
}

#include <stdio.h>

int main() {
    int a = 11, b = 9, carry;

    while (b) {

        // Carry is AND of a and b
        carry = a & b;

        // Sum without carry is XOR of a and b
        a = a ^ b;

        // Carry is shifted by one so that it can be
        // added in the next iteration
        b = carry << 1;
    }

    printf("%d", a);
    return 0;
}

public class Main {
    public static void main(String[] args) {
        int a = 11, b = 9, carry;

        while (b!= 0) {

            // Carry is AND of a and b
            carry = a & b;

            // Sum without carry is XOR of a and b
            a = a ^ b;

            // Carry is shifted by one so that it can be
            // added in the next iteration
            b = carry << 1;
        }

        System.out.println(a);
    }
}

def main():
    a = 11
    b = 9
    carry = 0

    while b!= 0:

        # Carry is AND of a and b
        carry = a & b

        # Sum without carry is XOR of a and b
        a = a ^ b

        # Carry is shifted by one so that it can be
        # added in the next iteration
        b = carry << 1

    print(a)

if __name__ == "__main__":
    main()

using System;

class Program {
    static void Main() {
        int a = 11, b = 9, carry;

        while (b!= 0) {

            // Carry is AND of a and b
            carry = a & b;

            // Sum without carry is XOR of a and b
            a = a ^ b;

            // Carry is shifted by one so that it can be
            // added in the next iteration
            b = carry << 1;
        }

        Console.WriteLine(a);
    }
}

function main() {
    let a = 11, b = 9, carry;

    while (b!= 0) {

        // Carry is AND of a and b
        carry = a & b;

        // Sum without carry is XOR of a and b
        a = a ^ b;

        // Carry is shifted by one so that it can be
        // added in the next iteration
        b = carry << 1;
    }

    console.log(a);
}

main();

20

Explanation: The algorithm repeatedly computes the sum without carry using XOR and the carry using AND until the carry becomes zero.