两个整数的按位递归加法

用手将两个二进制数相加时，请牢记进位并将其同时相加。但是要在程序中执行相同的操作，我们需要进行大量检查。递归解可以想象为进位和a ^ b (两个输入)的加法，直到进位变为0为止。

例子 ：

Input : int x = 45, y = 45
Output : 90

Input : int x = 4, y = 78
Output : 82

通过对两个位执行XOR(^)，可以获得两个位的总和。进位可以通过对两个位进行“与”(＆)获得。
上面是简单的Half Adder逻辑，可用于将2个单个位相加。我们可以将此逻辑扩展为整数。如果x和y在同一位置没有设置位，则x和y的按位XOR(^)得出x和y的总和。为了并入公共置位位，还使用了按位与(＆)。 x和y的按位与运算得出所有进位位。我们计算(x＆y)<< 1并将其添加到x ^ y中以获得所需的结果。

一个重要的观察结果是，如果(x＆y)变为0，则结果为x ^ y。

C

// CPP program to do recursive addition
// of two integers
#include 
  
int add(int x, int y) { 
    int keep = (x & y) << 1;
    int res = x^y;
  
    // If bitwise & is 0, then there
    // is not going to be any carry. 
    // Hence result of XOR is addition.
    if (keep == 0)
        return res;
  
    add(keep, res);
}
  
// Driver code
int main(){
    printf("%d", add(15, 38));
    return 0;
}

Java

// Java program to do recursive addition
// of two integers
import java.io.*;
  
class GFG {
      
    static int add(int x, int y)
    { 
        int keep = (x & y) << 1;
        int res = x^y;
      
        // If bitwise & is 0, then there
        // is not going to be any carry. 
        // Hence result of XOR is addition.
        if (keep == 0)
            return res;
              
        return add(keep, res);
    }
  
    // Driver code
    public static void main (String[] args)
    {
        System.out.println(add(15, 38));
    }
}
  
// This code is contributed by Ajit.

Python3

# Python program to do recursive addition
# of two integers
   
def add(x, y): 
    keep = (x & y) << 1;
    res = x^y;
   
    # If bitwise & is 0, then there
    # is not going to be any carry. 
    # Hence result of XOR is addition.
    if (keep == 0):
        return res;
   
    return add(keep, res);
  
   
# Driver code
  
print(add(15, 38));
  
#  This code is contributed by Princi Singh

C#

// C# program to do recursive
// addition of two integers
using System;
  
class GFG {
      
    static int add(int x, int y)
    { 
        int keep = (x & y) << 1;
        int res = x^y;
      
        // If bitwise & is 0, then there
        // is not going to be any carry. 
        // Hence result of XOR is addition.
        if (keep == 0)
            return res;
              
        return add(keep, res);
    }
  
    // Driver code
    public static void Main ()
    {
        Console.Write(add(15, 38));
    }
}
  
// This code is contributed by Smitha.

输出：

53