C中任何整数的旋转位 [英] Rotating bits of any integer in C

问题描述

将一个整数 2 传递给该函数，然后返回一个整数 4

Pass a integer 2 to this function and then return a integer which is 4

x = 2;
x = rotateInt('L', x, 1); 

(左移1位)

示例:00000010 -> 向左旋转 1 -> 00000100

Example: 00000010 -> rotate left by 1 -> 00000100

但如果我通过了:

x = rotateInt('R', x, 3); 

它将返回 64, 01000000

it will return 64, 01000000

这是代码，谁能纠正错误...谢谢

Here is the code, can someone correct the error... thanks

int rotateInt(char direction, unsigned int x, int y)
{
    unsigned int mask = 0;
    int num = 0, result = 0;
    int i;

    for (i = 0; i < y; i++)
    {     
        if (direction == 'R')
        {
            if ((x & 1) == 1)     
                x = (x ^ 129);
            else    
                x = x >> 1;
        }
        else if (direction == 'L')
        {
            if ((x & 128) == 1)  
                x = (x ^ 129);   
            else
                x = x << 1;
        }
    }
result = (result ^ x);
return result;   
}

推荐答案

所以，我假设你知道什么是右移和左移.而且您知道算术移位和逻辑移位之间的区别.

So, I'll assume you know what right and left shifts are. And that you know the difference between arithmetic and logical shifts.

C 只有算术移位.它不做逻辑移位，也不做旋转.好吧，我撒谎了，C 对无符号整数进行逻辑移位.

C only has arithmetic shifts. It doesn't do logical shifts, nor does it do rotates. okay I lied, C does logical shifts on unsigned ints.

旋转确实如此:它与逻辑移位相同，除了当您移动超过数字的末尾时，数字环绕"到另一侧.例如

A rotate does, well, exactly that: it's the same as a logical shift, except when you shift past the end of the number, the digits "wrap around" to the other side. For example

0010 右旋为 0001.如果你再次右转，你会得到 1000.看，1 环绕或旋转到整数的另一侧.

0010 right-rotated is 0001. If you right-rotate again, you get 1000. See, the 1 wrapped around, or rotated, to the other side of the integer.

左旋类似:0100左旋1000左旋0001左旋0010等

The left rotate is similar: 0100 left rotate 1000 left rotate 0001 left rotate 0010 etc.

请注意，旋转不会像算术右移那样保留符号位.

Note that rotates don't keep the sign bit as an arithmetic right-shift would.

所以，C 只有算术移位.所以你必须手动实现旋转"部分.所以，左转.你会想要:

So, C only has arithmetic shifts. So you have to implement the "rotate" part manually. So, take a left-rotate. You would want to:

1. 捕获最左边位的值.(是 0 还是 1?)
2. 左移
3. 根据我们从第 1 步捕获的内容，将最右边的位 - 我们在第 1 步中讨论的位(需要旋转)设置为正确的值.

您应该能够找出类似的右旋方法.

You should be able to figure out a similar method for right rotates.

祝你好运！

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