Para encontrar la suma de dos números sin usar ningún operador

Escriba un programa para encontrar la suma de enteros positivos sin usar ningún operador. Solo se permite el uso de printf(). No se puede utilizar ninguna otra función de biblioteca.

Solución 
Es una pregunta capciosa. Podemos usar printf() para encontrar la suma de dos números ya que printf() devuelve el número de caracteres impresos. El ancho del campo en printf()se puede utilizar para encontrar la suma de dos números. Podemos usar ‘*’ que indica el ancho mínimo de salida. Por ejemplo, en la sentencia “printf(“%*d”, ancho, num);”, el ‘ancho’ especificado se sustituye en lugar de *, y ‘num’ se imprime dentro del ancho mínimo especificado. Si el número de dígitos en ‘num’ es menor que el ‘ancho’ especificado, la salida se rellena con espacios en blanco. Si el número de dígitos es mayor, la salida se imprime tal cual (no truncada). En el siguiente programa, add() devuelve la suma de x e y. Imprime 2 espacios dentro del ancho especificado usando x e y. Entonces, el total de caracteres impresos es igual a la suma de x e y. Es por eso que add() devuelve x+y.

#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;
}

# 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.

#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;
}

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 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.

<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>

#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 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 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.

<?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>