To find sum of two numbers without using any operator

30 May, 2022
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++
C
Python3
#include <iostream>using namespace std;  int add(int x, int y){    return printf("%*c%*c", x, ' ', y, ' ');} // Driver codeint 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 codeint main(){    printf("Sum = %d", add(3, 4));    return 0;}
# Python code for the above approachdef add(x, y) :         return (x + y);    # Driver codeif __name__ == "__main__":         print("Sum = ", add(3, 4))     # This code is contributed by splvel62.
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++
C
Java
Python3
C#
Javascript
#include <iostream>using namespace std; int add(int x, int y){    return printf("%*c%*c", x, '\r', y, '\r');} // Driver codeint 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 codeint main(){    printf("Sum = %d", add(3, 4));    return 0;}
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 program for the above approachdef add(x, y) :         return (x + y);      # driver codeprint("Sum = ", add(3, 4)); # This code is contributed by sanjoy_62
// C# program for the above approachusing 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.
<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++
C
Java
Python3
C#
PHP
Javascript
#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 codeclass 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 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# codeusing 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 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.?>
<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)
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