在不使用任何运算符的情况下找到两个数字的和

编写一个程序来查找正整数之和，而无需使用任何运算符。只允许使用printf()。不能使用其他库函数。
解决方案
这是一个技巧问题。我们可以使用printf()查找两个数字的和，因为printf()返回打印的字符数。 printf()中的width字段可用于查找两个数字的和。我们可以使用“ *”来表示输出的最小宽度。例如，在语句“ printf(“％* d”，width，num);”中，指定的“ width”代替*，并且在指定的最小宽度内打印“ num”。如果’num’中的位数小于指定的’width’，则在输出中填充空格。如果位数更多，则按原样打印输出(不截断)。在下面的程序中，add()返回x和y的和。它在使用x和y指定的宽度内打印2个空格。因此，打印的总字符等于x和y的总和。这就是为什么add()返回x + y的原因。

C

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

输出：

Sum = 7

输出为七个空格，后跟“ Sum = 7”。我们可以通过使用回车符来避免前导空格。感谢krazyCoder和Sandeep提出的建议。以下程序打印输出，不带任何前导空格。

C

#include 
 
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;
}

输出：

Sum = 7

另一种方法：

C++

#include 
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 
 
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

的Python 3

# 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.

的PHP

 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.
?>

Java脚本

 
// 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);
 

输出：

sum = 15

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