8 different ways to Add Two Numbers in C/C++

Given two numbers A and B, the task is to find the sum of the given two numbers.
Examples:

Input: A = 5, B = 6 
Output: sum = 11
Input: A = 4, B = 11 
Output: sum = 15  

Method 1 – using Addition Operator: Here simply use the addition operator between two numbers and print the sum of the number. 

sum = A + B 

Below is the implementation of the above approach: 

C++

// C++ program to add two number
// using addition operator
#include <iostream>
using namespace std;
// Function to return sum
// of two number
int addTwoNumber(int A, int B)
{
    // Return sum of A and B
    return A + B;
}

// Driver Code
int main()
{
    // Given two number
    int A = 4, B = 11;

    // Function call
    cout << "sum = " << addTwoNumber(A, B);
    return 0;
}

Output

sum = 15

Time complexity: O(1)

Auxiliary Space: O(1)

Method 2 – using Subtraction Operator: Here simply use the subtraction operator between two numbers, two times so that minus-minus multiply and produce + operator and return the sum of the number.  

sum = A – (-B) 

Below is the implementation of the above approach:

C++

// C++ program to add two number
// using subtraction operator
#include <iostream>
using namespace std;
// Function to return sum
// of two number
int addTwoNumber(int A, int B)
{
    // Return sum of A and B
    return A - (-B);
}

// Driver Code
int main()
{
    // Given two number
    int A = 4, B = 11;

    // Function call
    cout << "sum = " << addTwoNumber(A, B);
    return 0;
}

Output

sum = 15

Time complexity: O(1)

Auxiliary Space: O(1)

Method 3 – using increment/decrement operator: Here use increment/decrement operator while one number decrement to zero and in another number increment by one when the first number decrement by one, return the second number. 

while(B > 0) { 
A++; 
B–; 
}  

Below is the implementation of the above approach: 

C++

// C++ program to add two number using
// increment/decrement operator
#include <iostream>
using namespace std;

// Function to return sum
// of two number
int addTwoNumber(int A, int B)
{
    // When A is positive
    while (A > 0) {
        A--;
        B++;
    }

    // When A is negative
    while (A < 0) {
        A++;
        B--;
    }

    // Return sum of A and B
    return B;
}

// Driver Code
int main()
{
    // Given two number
    int A = 4, B = 11;

    // Function call
    cout << "sum = " << addTwoNumber(A, B);
    return 0;
}

Output

sum = 15

Time Complexity: O(N) where N is the min(A,B)
Auxiliary Space: O(1)

Method 4 – using printf() method: Here “%*s” specifier print value of a variable, the value of variable times, and printf return how many character print on the screen.

printf(“%*s%*s”, A, “”, B, “”); 

Below is the implementation of the above approach: 

C++

// C++ program to add two number
// using printf method
#include <iostream>
using namespace std;
// Function to return sum
// of two number
int addTwoNumber(int A, int B)
{
    // Return sum of A and B
    return printf("%*s%*s", A, "", B, "");
}

// Driver Code
int main()
{
    // Given two number
    int A = 4, B = 11;

    // Function call
    printf("sum = %d", addTwoNumber(A, B));
    return 0;
}

Output

               sum = 15

Time complexity: O(1)

Auxiliary Space: O(1)

Method 5 – using Half Adder method: A sum of two bits can be obtained by performing Bitwise XOR(^) of the two bits. Carry bit can be obtained by performing Bitwise AND(&) of two bits. 
Above is simple Half Adder logic that can be used to add 2 single bits. We can extend this logic for integers. If x and y don’t have set bits at the same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. 

Sum = A & B; 
Carry = x ^ y 

Below is the implementation of the above approach: 

C++

// C++ program to add two number
// using half adder method
#include <iostream>
using namespace std;

// Function to return sum
// of two number
int addTwoNumber(int A, int B)
{

    // Iterate till there is no carry
    while (B != 0) {

        // Carry now contains common
        // set bits of A and B
        int carry = A & B;

        // Sum of bits of A and B
        // where at least one of the
        // bits is not set
        A = A ^ B;

        // Carry is shifted by one so
        // that adding it to A gives
        // the required sum
        B = carry << 1;
    }

    return A;
}

// Driver Code
int main()
{
    // Given two number
    int A = 4, B = 11;

    // Function call
    printf("sum = %d", addTwoNumber(A, B));
    return 0;
}

Output

sum = 15

Time complexity: O(logn)

Auxiliary Space: O(1)

Method 6 – using exponential and logarithm: The idea is to find the exponential of the given two numbers and print the logarithmic of the result.

printf(“%g\n”, log(exp(A) * exp(B))); 

Below is the implementation of the above approach: 

C++

// C++ program to add two number
// using log and exponential
#include <bits/stdc++.h>
using namespace std;

// Function to return sum
// of two number
int addTwoNumber(int A, int B)
{
    // Return sum of A and B
    return log(exp(A) * exp(B));
}

// Driver Code
int main()
{
    // Given two number
    int A = 4, B = 11;

    // Function call
    printf("sum = %d", addTwoNumber(A, B));
    return 0;
}

Output

sum = 15

Time complexity: O(logn)

Auxiliary Space: O(1)

Method 7 – using Recursion:

1. Get the numbers A and B whose sum is to be calculated.
2. Base Case: If A is greater than 0, then return B.

if(A > 0) {
   return B;
}

3. Recursive Call: Update A to (A&B)<<1 and B to A ^ B and recursively call for the updated value. 

recursive_function((A & B) << 1, A ^ B);

Below is the implementation of the above approach:

C++

// C++ program to add two number
// using Recursion
#include <iostream>

// Function to return sum
// of two number
int addTwoNumber(int A, int B)
{
    // Base Case
    if (!A)
        return B;

    // Recursive Call
    else
        return addTwoNumber((A & B) << 1, A ^ B);
}

// Driver Code
int main()
{
    // Given two number
    int A = 4, B = 11;

    // Function call
    printf("sum = %d", addTwoNumber(A, B));
    return 0;
}

Output

sum = 15

Time complexity: O(logn)

Auxiliary Space: O(logn)

Method 8 –  using this pointer

Taking two numbers x and y from the user and the task is to find the sum of the given two numbers using this pointer.

Example:

input:

Enter Two Number: 5

4

output:

Sum is: 9

This can be done by accessing the data member of an object by using this pointer and performing an addition operation between them.

C++

// CPP program for above approach
#include <iostream>
using namespace std;

class A
{
    int a, b, sum;

public:
    A(int x, int y)
    {
        a = x;
        b = y;
    }

    // Using this pointer to
    // access variable
    void calcSum()
    {
      sum = this->a + this->b;
    }

    void showSum()
    {
      cout << "Sum is: " << sum << endl;
    }
};

// Driver Code
int main()
{
    int x, y;
    cout << "Enter Two Numbers: ";
    x = 4, y = 5;
    A a(x, y);
    a.calcSum();
    a.showSum();
    return 0;
}

Output

Enter Two Numbers: Sum is: 9

Time complexity: O(1)

Auxiliary Space: O(1)