使用按位运算符改变符号 [英] Change sign using bitwise operators

问题描述

如何更改INT使用位运算符的标志？显然，我们可以使用 X * = - 1 或 X / = - 1 。是否有这样做的任何最快的方法？

我做了如下小的考验。只是出于好奇...

 公共类ChangeSign {
    公共静态无效的主要（字串[] args）{
        INT X = 198347;
        INT回路= 1000000;
        诠释Ÿ;
        长启动= System.nanoTime（）;
        的for（int i = 0; I＆LT; LOOP;我++）{
            Y =（〜X）+ 1;
        }
        长MID1 = System.nanoTime（）;
        的for（int i = 0; I＆LT; LOOP;我++）{
            Y = -x;
        }
        长MID2 = System.nanoTime（）;
        的for（int i = 0; I＆LT; LOOP;我++）{
            Y = X * -1;
        }
        长MID3 = System.nanoTime（）;
        的for（int i = 0; I＆LT; LOOP;我++）{
            Y = X / -1;
        }
        长端= System.nanoTime（）;
        的System.out.println（MID1  - 启动）;
        的System.out.println（MID2  -  MID1）;
        的System.out.println（MID3  -  MID2）;
        的System.out.println（完 -  MID3）;
    }
}
 

输出几乎类似于：

  2200211
835772
1255797
4651923
 

解决方案

非浮点之间的速度差（例如INT数学）加法/乘法和位运算小于上可以忽略不计，几乎所有的机器。

有仅使用按位操作来打开一个n位符号整数到其负当量，作为反运算看起来像 X =（〜X）+ 1 ，这需要一个补充。然而，假设有符号整数为32位，你可能可以写一个方程式逐位做到这一点的计算。 注意：不这样做。

否定了一些最常见的，易读的方式是 X = -x 。

How to change the sign of int using bitwise operators? Obviously we can use x*=-1 or x/=-1. Is there any fastest way of doing this?

I did a small test as below. Just for curiosity...

public class ChangeSign {
    public static void main(String[] args) {
        int x = 198347;
        int LOOP = 1000000;
        int y;
        long start = System.nanoTime();
        for (int i = 0; i < LOOP; i++) {
            y = (~x) + 1;
        }
        long mid1 = System.nanoTime();
        for (int i = 0; i < LOOP; i++) {
            y = -x;
        }
        long mid2 = System.nanoTime();
        for (int i = 0; i < LOOP; i++) {
            y = x * -1;
        }
        long mid3 = System.nanoTime();
        for (int i = 0; i < LOOP; i++) {
            y = x / -1;
        }
        long end = System.nanoTime();
        System.out.println(mid1 - start);
        System.out.println(mid2 - mid1);
        System.out.println(mid3 - mid2);
        System.out.println(end - mid3);
    }
}

Output is almost similar to :

2200211
835772
1255797
4651923

解决方案

The speed difference between non-floating point (e.g. int math) addition/multiplication and bitwise operations is less than negligible on almost all machines.

There is no general way to turn an n-bit signed integer into its negative equivalent using only bitwise operations, as the negation operation looks like x = (~x) + 1, which requires one addition. However, assuming the signed integer is 32 bit you can probably write a bitwise equation to do this calculation. Note: do not do this.

The most common, readable way to negate a number is x = -x.

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