Question

我目前正在使用BitConverter在signed int中打包两个unsigned short。此代码对不同的值执行数百万次，我认为代码可以进一步优化。这是我目前正在做的事情 - 您可以假设代码是C＃/ NET。

// to two unsigned shorts from one signed int:
int xy = 343423;
byte[] bytes = BitConverter.GetBytes(xy);
ushort m_X = BitConverter.ToUInt16(bytes, 0);
ushort m_Y = BitConverter.ToUInt16(bytes, 2);

// convet two unsigned shorts to one signed int
byte[] xBytes = BitConverter.GetBytes(m_X);
byte[] yBytes = BitConverter.GetBytes(m_Y);
byte[] bytes = new byte[] {
   xBytes[0],
   xBytes[1],
   yBytes[0],
   yBytes[1],
 };
 return BitConverter.ToInt32(bytes, 0);

所以我发现如果我进行bithift，我可以避免构造数组的开销。但是对于我的生活，我无法弄清楚正确的换档操作是什么。我的第一次可悲的尝试涉及以下代码：

int xy = 343423;
const int mask = 0x00000000;
byte b1, b2, b3, b4;
b1 = (byte)((xy >> 24));
b2 = (byte)((xy >> 16));
b3 = (byte)((xy >> 8) & mask);
b4 = (byte)(xy & mask);
ushort m_X = (ushort)((xy << b4) | (xy << b3));
ushort m_Y = (ushort)((xy << b2) | (xy << b1));

有人可以帮助我吗？我想我需要在移位前屏蔽上下字节。我看到的一些例子包括使用type.MaxValue或任意数字减去，比如负12，这非常令人困惑。

**更新**

谢谢你的答案。以下是基准测试的结果：

// 34ms for bit shift with 10M operations
// 959ms for BitConverter with 10M operations

static void Main(string[] args)
    {
        Stopwatch stopWatch = new Stopwatch();

        stopWatch.Start();
        for (int i = 0; i < 10000000; i++)
        {
            ushort x = (ushort)i;
            ushort y = (ushort)(i >> 16);
            int result = (y << 16) | x;
        }
        stopWatch.Stop();
        Console.WriteLine((int)stopWatch.Elapsed.TotalMilliseconds + "ms");

        stopWatch.Start();
        for (int i = 0; i < 10000000; i++)
        {
            byte[] bytes = BitConverter.GetBytes(i);
            ushort x = BitConverter.ToUInt16(bytes, 0);
            ushort y = BitConverter.ToUInt16(bytes, 2);

            byte[] xBytes = BitConverter.GetBytes(x);
            byte[] yBytes = BitConverter.GetBytes(y);
            bytes = new byte[] {
                xBytes[0],
                xBytes[1],
                yBytes[0],
                yBytes[1],
            };
            int result = BitConverter.ToInt32(bytes, 0);
        }
        stopWatch.Stop();
        Console.WriteLine((int)stopWatch.Elapsed.TotalMilliseconds + "ms");


        Console.ReadKey();
    }

Answer 1

最简单的方法是使用两个班次：

int xy = -123456;
// Split...
ushort m_X = (ushort) xy;
ushort m_Y = (ushort)(xy>>16);
// Convert back...
int back = (m_Y << 16) | m_X;

关于ideone的演示：link。

Answer 2

int xy = 343423;
ushort low = (ushort)(xy & 0x0000ffff);
ushort high = (ushort)((xy & 0xffff0000) >> 16);
int xxyy = low + (((int)high) << 16);

将signed int转换为两个unsigned short以进行重构

2 个答案: