所以我有这个算法，它左右旋转

SO I have this algorithm that does left rotation

    public static ushort RotateLeft(ushort value, int count)
    {
        int left = value << count;
        int right = value >> (16 - count);
        return (ushort)(left | right);
    }

然而，这是不生产我想例如价值观，如果我有18的值时，左移1位它的结果应该是3，而是这只是增加了一个零结尾：

However this is not producing the values I want for example if I have a value of 18 when rotated left by 1 bit it the result should be 3 but instead this just adds a zero at the end:

这是什么样的算法做：

10010     18
100100    36

这就是我想要的：

10010     18
00101      3

该位应该被移出的符号位的位置（位0）进入至少显著比特位置（位15），以及，因此，没有比特被丢失。

The bits should be shifted out of the sign bit position (bit 0) enter the least significant bit position (bit 15) and, consequently, no bits are lost.

我想在这里说明什么： http://www.xgc.com/manuals/m1750-ada/m1750 /x2733.html

What I want is described here: http://www.xgc.com/manuals/m1750-ada/m1750/x2733.html

这是一个CRC算法，我想转换为C＃

It's a CRC algorithm that I want to convert to C#

推荐答案

您code应该作如下更正。

Your code should be corrected as follows.

我添加了新的说法numberOfBits来位被旋转了presents数量。

I add new argument 'numberOfBits' to presents number of bits to be rotated.

public static ushort RotateLeft(ushort value, int numberOfBits int countToRotate)
{
    countToRotate = countToRotate mod numberOfBits; // in case of rotate more than once.
    ushort mask = -1  // 1s for all 16 bits.
    mask = mask << numberOfBits;

    int left = value << countToRotate;  // rotate left
    int right = left >> (numberOfBits); // move left-overflowed bits to right
    return (ushort)((left | right) & mask);
}

这code正常工作，如果（numOfBits + countToRotate）LT; = 16

This code works correctly if (numOfBits+countToRotate)<=16.