不使用算术运算运算符将两个数字相加

编写一个函数Add()返回两个整数的和。该函数不应使用任何算术运算运算符(+，++，–，-，..等)。
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。

C++

// C++ Program to add two numbers
// without using arithmetic operator
#include 
using namespace std;
 
int Add(int x, int y)
{
    // Iterate till there is no carry
    while (y != 0)
    {
        // carry now contains common
        //set bits of x and y
        int carry = x & y;
 
        // Sum of bits of x and y where at
        //least one of the bits is not set
        x = x ^ y;
 
        // Carry is shifted by one so that adding
        // it to x gives the required sum
        y = carry << 1;
    }
    return x;
}
 
// Driver code
int main()
{
    cout << Add(15, 32);
    return 0;
}
 
// This code is contributed by rathbhupendra

C

// C Program to add two numbers
// without using arithmetic operator
#include
 
int Add(int x, int y)
{
    // Iterate till there is no carry 
    while (y != 0)
    {
        // carry now contains common
        //set bits of x and y
        int carry = x & y; 
 
        // Sum of bits of x and y where at
        //least one of the bits is not set
        x = x ^ y;
 
        // Carry is shifted by one so that adding
        // it to x gives the required sum
        y = carry << 1;
    }
    return x;
}
 
int main()
{
    printf("%d", Add(15, 32));
    return 0;
}

Java

// Java Program to add two numbers
// without using arithmetic operator
import java.io.*;
 
class GFG
{
    static int Add(int x, int y)
    {
        // Iterate till there is no carry
        while (y != 0)
        {
            // carry now contains common
            // set bits of x and y
            int carry = x & y;
 
            // Sum of bits of x and
            // y where at least one
            // of the bits is not set
            x = x ^ y;
 
            // Carry is shifted by
            // one so that adding it
            // to x gives the required sum
            y = carry << 1;
        }
        return x;
    }
     
    // Driver code
    public static void main(String arg[])
    {
        System.out.println(Add(15, 32));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 Program to add two numbers
# without using arithmetic operator
def Add(x, y):
 
    # Iterate till there is no carry
    while (y != 0):
     
        # carry now contains common
        # set bits of x and y
        carry = x & y
 
        # Sum of bits of x and y where at
        # least one of the bits is not set
        x = x ^ y
 
        # Carry is shifted by one so that  
        # adding it to x gives the required sum
        y = carry << 1
     
    return x
 
print(Add(15, 32))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# Program to add two numbers
// without using arithmetic operator
using System;
 
class GFG
{
    static int Add(int x, int y)
    {
        // Iterate till there is no carry
        while (y != 0)
        {
            // carry now contains common
            // set bits of x and y
            int carry = x & y;
 
            // Sum of bits of x and
            // y where at least one
            // of the bits is not set
            x = x ^ y;
 
            // Carry is shifted by
            // one so that adding it
            // to x gives the required sum
            y = carry << 1;
        }
        return x;
    }
     
    // Driver code
    public static void Main()
    {
        Console.WriteLine(Add(15, 32));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Javascript Program to add two numbers
// without using arithmetic operator
 
   
    function Add(x, y) {
        // Iterate till there is no carry  
        while (y != 0)
        {
            // carry now contains common 
            //set bits of x and y
            let carry = x & y;  
   
            // Sum of bits of x and y where at 
            //least one of the bits is not set
            x = x ^ y; 
   
            // Carry is shifted by one so that adding
            // it to x gives the required sum
            y = carry << 1;
        }
        return x;
    }
   
     //driver code
    document.write(Add(15, 32));
  
// This code is contributed by Surbhi Tyagi

输出 ：

47

以下是相同方法的递归实现。

C

int Add(int x, int y)
{
    if (y == 0)
        return x;
    else
        return Add( x ^ y, (x & y) << 1);
}

Java

static int Add(int x, int y)
{
  if (y == 0)
    return x;
  else
    return Add(x ^ y, (x & y) << 1);
}
 
// This code is contributed by subham348

Python3

def Add(x, y):
   
    if (y == 0):
        return x
    else
        return Add( x ^ y, (x & y) << 1)
     
# This code is contributed by subhammahato348

C＃

static int Add(int x, int y)
{
  if (y == 0)
    return x;
  else
    return Add(x ^ y, (x & y) << 1);
}
 
// This code is contributed by subhammahato348

Java脚本

function Add(x, y)
{
    if (y == 0)
        return x;
    else
        return Add(x ^ y, (x & y) << 1);
}
 
// This code is contributed by Ankita saini