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To find sum of two numbers without using any operator

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Write a program to find sum of positive integers without using any operator. Only use of printf() is allowed. No other library function can be used.

Solution 
It’s a trick question. We can use printf() to find sum of two numbers as printf() returns the number of characters printed. The width field in printf() can be used to find the sum of two numbers. We can use ‘*’ which indicates the minimum width of output. For example, in the statement “printf(“%*d”, width, num);”, the specified ‘width’ is substituted in place of *, and ‘num’ is printed within the minimum width specified. If number of digits in ‘num’ is smaller than the specified ‘width’, the output is padded with blank spaces. If number of digits are more, the output is printed as it is (not truncated). In the following program, add() returns sum of x and y. It prints 2 spaces within the width specified using x and y. So total characters printed is equal to sum of x and y. That is why add() returns x+y.

C++




#include <iostream>
using namespace std;
  
int add(int x, int y)
{
    return printf("%*c%*c", x, ' ', y, ' ');
}
 
// Driver code
int main()
{
    printf("Sum = %d", add(3, 4));
    return 0;
}
 
// This code is contributed by shubhamsingh10

C




#include <stdio.h>
 
int add(int x, int y)
{
    return printf("%*c%*c", x, ' ', y, ' ');
}
 
// Driver code
int main()
{
    printf("Sum = %d", add(3, 4));
    return 0;
}

Java




public class Add
{
    public static void main(String[] args)
    {
        int x = 3, y = 4;
        System.out.print("Sum = " + add(x, y));
    }
  
    public static int add(int x, int y)
    {
        String str = String.format("%" + x + "c%" + y + "c", ' ', ' ');
        return str.length();
    }
}

Python3




# Python code for the above approach
def add(x, y) :
     
    return (x + y);
    
# Driver code
if __name__ == "__main__":
     
    print("Sum = ", add(3, 4))
 
    # This code is contributed by splvel62.

C#




using System;
 
namespace Add
{
    class Program
    {
        static void Main(string[] args)
        {
            int x = 3, y = 4;
            Console.WriteLine("Sum = " + Add(x, y));
        }
 
        static int Add(int x, int y)
        {
            string str = string.Format("{0," + x + "}{1," + y + "}", ' ', ' ');
            return str.Length;
        }
    }
}

Javascript




function add(x, y)
{
    return console.log("%*c%*c", x, ' ', y, ' ');
}
 
// Driver code
function main()
{
    console.log("Sum = %d", add(3, 4));
    return 0;
}
 
main();
 
// This code is contributed by factworx412

Output: 

Sum = 7

Time Complexity: O(1)
Auxiliary Space: O(1)

The output is seven spaces followed by “Sum = 7”. We can avoid the leading spaces by using carriage return. Thanks to krazyCoder and Sandeep for suggesting this. The following program prints output without any leading spaces.

C++




#include <iostream>
using namespace std;
 
int add(int x, int y)
{
    return printf("%*c%*c", x, '\r', y, '\r');
}
 
// Driver code
int main()
{
    printf("Sum = %d", add(3, 4));
    return 0;
}
 
// This code is contributed by shubhamsingh10

C




#include <stdio.h>
 
int add(int x, int y)
{
    return printf("%*c%*c", x, '\r', y, '\r');
}
 
// Driver code
int main()
{
    printf("Sum = %d", add(3, 4));
    return 0;
}

Java




class GFG {
 
    static int add(int x, int y) {
        return (x + y);
    }
 
    // Driver code
    public static void main(String[] args) {
        System.out.printf("Sum = %d", add(3, 4));
    }
}
 
// This code is contributed by Rajput-Ji

Python3




# Python program for the above approach
def add(x, y) :
     
    return (x + y);
     
 
# driver code
print("Sum = ", add(3, 4));
 
# This code is contributed by sanjoy_62

C#




// C# program for the above approach
using System;
 
public class GFG {
 
  static int add(int x, int y)
  {
    return (x + y);
  }
 
 
  // Driver Code
  public static void Main(String[] args) {
 
    Console.WriteLine("Sum = " + add(3, 4));
  }
}
 
// This code is contributed by code_hunt.

Javascript




<script>
    // JavaScript code for the above approach
 
  function add(x, y)
  {
    return (x + y);
  }
 
    // Driver Code
    document.write("Sum = " + add(3, 4));
     
    // This code is contributed by avijitmondal1998.
</script>

Output: 

      Sum = 7

Time Complexity: O(1)

Auxiliary Space: O(1)

Another Method : 

C++




#include <iostream>
using namespace std;
 
int main()
{
    int a = 10, b = 5;
    if (b > 0) {
        while (b > 0) {
            a++;
            b--;
        }
    }
    if (b < 0) { // when 'b' is negative
        while (b < 0) {
            a--;
            b++;
        }
    }
    cout << "Sum = " << a;
    return 0;
}
 
// This code is contributed by SHUBHAMSINGH10
// This code is improved & fixed by Abhijeet Soni.

C




#include <stdio.h>
 
int main()
{
    int a = 10, b = 5;
    if (b > 0) {
        while (b > 0) {
            a++;
            b--;
        }
    }
    if (b < 0) { // when 'b' is negative
        while (b < 0) {
            a--;
            b++;
        }
    }
    printf("Sum = %d", a);
    return 0;
}
 
// This code is contributed by Abhijeet Soni

Java




// Java code
class GfG {
 
    public static void main(String[] args)
    {
        int a = 10, b = 5;
        if (b > 0) {
            while (b > 0) {
                a++;
                b--;
            }
        }
        if (b < 0) { // when 'b' is negative
            while (b < 0) {
                a--;
                b++;
            }
        }
        System.out.println("Sum is: " + a);
    }
}
 
// This code is contributed by Abhijeet Soni

Python3




# Python 3 Code
 
if __name__ == '__main__':
     
    a = 10
    b = 5
 
    if b > 0:
        while b > 0:
            a = a + 1
            b = b - 1
    if b < 0:
        while b < 0:
            a = a - 1
            b = b + 1
     
    print("Sum is: ", a)
 
# This code is contributed by Akanksha Rai
# This code is improved & fixed by Abhijeet Soni

C#




// C# code
using System;
 
class GFG {
    static public void Main()
    {
        int a = 10, b = 5;
        if (b > 0) {
            while (b > 0) {
                a++;
                b--;
            }
        }
        if (b < 0) { // when 'b' is negative
            while (b < 0) {
                a--;
                b++;
            }
        }
        Console.Write("Sum is: " + a);
    }
}
 
// This code is contributed by Tushil
// This code is improved & fixed by Abhijeet Soni.

PHP




<?php
// PHP Code
$a = 10;
$b = 5;
 
if ($b > 0) {
while($b > 0)
{
    $a++;
    $b--;
}
}
 
if ($b < 0) {
while($b < 0)
{
    $a--;
    $b++;
}
}
 
 
echo "Sum is: ", $a;
 
// This code is contributed by Dinesh
// This code is improved & fixed by Abhijeet Soni.
?>

Javascript




<script>
 
// Javascript program for the above approach
 
// Driver Code
 
    let a = 10, b = 5;
    if (b > 0) {
        while (b > 0) {
            a++;
            b--;
        }
    }
    if (b < 0) { // when 'b' is negative
        while (b < 0) {
            a--;
            b++;
        }
    }
    document.write("Sum = " + a);
 
</script>

Output: 

sum = 15

Time Complexity: O(b)
Auxiliary Space: O(1)

Approach: Bit Manipulation

Steps:

  1. Calculate the sum of the numbers without taking into account the carry.
  2. Calculate the carry using bitwise AND and shift left operator.
  3. Add the carry to the sum.
  4. Repeat steps 1-3 until there is no carry left.

C++




// C++ program for the above approach
#include <iostream>
 
// Function to add two number without
// operators
int add_without_operator(int a, int b)
{
 
    while (b != 0) {
 
        // Calculate sum without carry
        int sum = a ^ b;
 
        // Calculate carry
        int carry = (a & b) << 1;
 
        // Add sum and carry
        a = sum;
        b = carry;
    }
 
    return a;
}
 
// Driver Code
int main()
{
    int a = 5;
    int b = 7;
    int result = add_without_operator(a, b);
    std::cout << "The sum of " << a << " and " << b
              << " is: " << result << std::endl;
 
    return 0;
}

Java




public class AddWithoutOperator {
    public static int add(int a, int b)
    {
        while (b != 0) {
            int sum = a ^ b; // XOR operation to calculate
                             // sum without carry
            int carry
                = (a & b)
                  << 1; // AND and left shift operation to
                        // calculate carry
            a = sum;
            b = carry;
        }
        return a;
    }
 
    public static void main(String[] args)
    {
        int a = 5;
        int b = 7;
        int result = add(a, b);
        System.out.println("The sum of " + a + " and " + b
                           + " is: " + result);
    }
}

Python3




def add_without_operator(a: int, b: int) -> int:
    while b != 0:
        # Calculate sum without carry
        sum = a ^ b
         
        # Calculate carry
        carry = (a & b) << 1
         
        # Add sum and carry
        a = sum
        b = carry
     
    return a
 
a = 5
b = 7
result = add_without_operator(a, b)
print(f"The sum of {a} and {b} is: {result}")

C#




using System;
 
class Program
{
    // Function to add two number without operators
    static int AddWithoutOperator(int a, int b)
    {
        while (b != 0)
        {
            // Calculate sum without carry
            int sum = a ^ b;
 
            // Calculate carry
            int carry = (a & b) << 1;
 
            // Add sum and carry
            a = sum;
            b = carry;
        }
 
        return a;
    }
 
    // Driver Code
    static void Main(string[] args)
    {
        int a = 5;
        int b = 7;
        int result = AddWithoutOperator(a, b);
        Console.WriteLine("The sum of {0} and {1} is: {2}", a, b, result);
    }
}

Javascript




function addWithoutOperator(a, b) {
while (b !== 0) {
// Calculate sum without carry
let sum = a ^ b;
// Calculate carry
let carry = (a & b) << 1;
 
// Add sum and carry
a = sum;
b = carry;
}
 
return a;
}
 
// Driver Code
let a = 5;
let b = 7;
let result = addWithoutOperator(a, b);
console.log(`The sum of ${a} and ${b} is: ${result}`);

Output

The sum of 5 and 7 is: 12

Time Complexity: O(log N), where N is the maximum number of bits in a or b.
Auxiliary Space: O(1)

Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above


Last Updated : 26 Apr, 2023
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