使用逐比特反转 [英] Bit Reversal using bitwise

问题描述

我试图做到位反转一个字节。我用下面

I am trying to do bit reversal in a byte. I use the code below

static int BitReversal(int n)
{
    int u0 = 0x55555555; // 01010101010101010101010101010101
    int u1 = 0x33333333; // 00110011001100110011001100110011
    int u2 = 0x0F0F0F0F; // 00001111000011110000111100001111
    int u3 = 0x00FF00FF; // 00000000111111110000000011111111
    int u4 = 0x0000FFFF;
    int x, y, z;
    x = n;
    y = (x >> 1) & u0;
    z = (x & u0) << 1;
    x = y | z;

    y = (x >> 2) & u1;
    z = (x & u1) << 2;
    x = y | z;

    y = (x >> 4) & u2;
    z = (x & u2) << 4;
    x = y | z;

    y = (x >> 8) & u3;
    z = (x & u3) << 8;
    x = y | z;

    y = (x >> 16) & u4;
    z = (x & u4) << 16;
    x = y | z;

    return x;
}

有可反向的位（32位的计算机上），但有一个问题， 例如，该输入是10001111101，我想获得10111110001，但这种方法会扭转整个字节包括标题0。输出为10111110001000000000000000000000。 有没有什么方法来仅扭转实际数量？我不希望将其转换为字符串和反向，然后再进行转换。有没有纯粹的数学方法或位操作方法？

It can reverser the bit (on a 32-bit machine), but there is a problem, For example, the input is 10001111101, I want to get 10111110001, but this method would reverse the whole byte including the heading 0s. The output is 10111110001000000000000000000000. Is there any method to only reverse the actual number? I do not want to convert it to string and reverser, then convert again. Is there any pure math method or bit operation method?

最好的问候，

推荐答案

不好的方法是转移，直到你得到一个1右侧：

Cheesy way is to shift until you get a 1 on the right:

if (x != 0) {
    while ((x & 1) == 0) {
        x >>= 1;
    }
}

请注意：您应该所有的变量切换到无符号整型。正如写你可以有多余的符号扩展任何时候你右移。

Note: You should switch all the variables to unsigned int. As written you can have unwanted sign-extension any time you right shift.

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