为什么另外这款code（使用位运算）的作品在Java中 [英] Why this code for addition(using bitwise operation) works in java

问题描述

public int add(int a, int b){  
        while (b != 0){
            int carry = (a & b) ;

            a = a ^ b; 

            b = carry << 1;
        }
        return a;
 }

这是code来计算使用按位运算两个整数的总和。

This is the code to calculate sum of two integers using bitwise operation.

如果我手动/编程计算，我看到它为每一个整数。
但我无法弄清楚 A 和执行的中间值之间的关系。
为什么进位乘以2指派给 B ？

If i calculate manually/programmatically, i see it works for every integer. But i am not able to figure out any relation between intermediate value of a and carry. and why carry is multiplied by 2 to assign to b?

PS：我找到了一个答案在这里
按位乘法和加法在Java中
但是这是乘法而不是加法

PS: I found the one answer here Bitwise Multiply and Add in Java but this is for multiplication and not for addition.

推荐答案

先来回顾一下除了小学。例如26 + 147 = 173.您可以通过6 + 7 = 13开始，所以你把3的总和，并携带一个，等等 - 那就是：增加了两个数字，如果有必要

First recall addition in primary school. e.g. 26 + 147 = 173. You start by 6+7=13, so you put 3 in the sum, and carry the one, and so forth - that is: you add two digits and carry a one if necessary.

carry:  1
a:      26
b:     147
-----------------
sum:   173

在code确实对二进制数几乎同样的事情，但有一个小调整。而不是采取一个数字位置的时间，它需要所有一气呵成。代替包括i。从位置i-1的进位（即，包括1加入2和4时），则code添加所有龋在第二次迭代。那么，它的作用是： 026 + 147 = 163 + 010 = 173 + 000

有关二进制数字的= 6 = 00110以及b = 7 = 00101你

For the binary numbers a=6=00110 and b=7=00101 you get

首先，你发现携带;这是所有职位，其中两个 A 和 B 有其位设置： INT携带=（ A和b）;

First you find the carries; that is all positions where both a and b has its bit set: int carry = (a & b) ;

然后ID不会在另外的数字，忽略进位，并将其存储 A ： A = A ^ B; 这将响应 6 + 7 = 3 中的例子。

Then id does the addition of digits, ignoring the carry, and stores it in a: a = a ^ b; This will respond to 6+7=3 in the example.

最后一部分转移进到下一个数字位置，即确保在本例中的1搭载的是感动，从1对10的：携带和LT;＆LT; 1;

The last part shifts the carry to the next digit position, i.e. ensuring the 1-carry in the example is "moved" from the 1's to the 10's: carry << 1;

while循环继续只要有携带尚未列入总和。

The while loop continues as long as there are carries that has not been included in the sum.

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