To find sum of two numbers without using any operator

Last Updated : 02 Jul, 2024

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

#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();     } }

Python

# 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.

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

#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

Python

# 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# 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.

#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

Python

# 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# 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.

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>

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. ?>

Output:

sum = 15

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

Approach: Bit Manipulation

Steps:

Calculate the sum of the numbers without taking into account the carry.
Calculate the carry using bitwise AND and shift left operator.
Add the carry to the sum.
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);     } }

Python

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}")

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}`);