我发现执行此操作最快的代码来自Bit Twiddling
Hacks网站。具体而言，基于DeBruijn序列的查找。参见http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogDeBruijn
我测试了一个朴素的方法，一个基于开关的方法以及两个Bit Twiddling
Hacks方法：DeBruijn序列，另一个表示“如果您知道自己的价值是2的幂”。
我将所有这些与3200万的2的幂进行比较。也就是说，形式为2 ^ N的整数，其中N在0..30的范围内。2 ^
31的值是负数，这会使幼稚的方法陷入无限循环。
我在发布模式下使用Visual Studio 2010编译了代码，然后在没有调试器（即Ctrl +
F5）的情况下运行了该代码。在我的系统上，数十次运行的平均值是：
- 天真的方法：950毫秒
- 切换方式：660毫秒
- Bithack方法1：1,154毫秒
- DeBruijn：105毫秒
显然，DeBruijn序列方法比任何其他方法都快得多。另一个Bithack方法在这里较差，因为从C到C＃的转换会导致效率低下。例如，C语句
int r =(v & b[0]) != 0;最终需要
if在C＃中使用或三元运算符（即？:）。
这是代码。
class Program{ const int Million = 1000 * 1000; static readonly int NumValues = 32 * Million; static void Main(string[] args) { // Construct a table of integers. // These are random powers of two. // That is 2^N, where N is in the range 0..31. Console.WriteLine("Constructing table"); int[] values = new int[NumValues]; Random rnd = new Random(); for (int i = 0; i < NumValues; ++i) { int pow = rnd.Next(31); values[i] = 1 << pow; } // Run each one once to make sure it's JITted GetLog2_Bithack(values[0]); GetLog2_DeBruijn(values[0]); GetLog2_Switch(values[0]); GetLog2_Naive(values[0]); Stopwatch sw = new Stopwatch(); Console.Write("GetLog2_Naive ... "); sw.Restart(); for (int i = 0; i < NumValues; ++i) { GetLog2_Naive(values[i]); } sw.Stop(); Console.WriteLine("{0:N0} ms", sw.ElapsedMilliseconds); Console.Write("GetLog2_Switch ... "); sw.Restart(); for (int i = 0; i < NumValues; ++i) { GetLog2_Switch(values[i]); } sw.Stop(); Console.WriteLine("{0:N0} ms", sw.ElapsedMilliseconds); Console.Write("GetLog2_Bithack ... "); sw.Restart(); for (int i = 0; i < NumValues; ++i) { GetLog2_Bithack(values[i]); } Console.WriteLine("{0:N0} ms", sw.ElapsedMilliseconds); Console.Write("GetLog2_DeBruijn ... "); sw.Restart(); for (int i = 0; i < NumValues; ++i) { GetLog2_DeBruijn(values[i]); } sw.Stop(); Console.WriteLine("{0:N0} ms", sw.ElapsedMilliseconds); Console.ReadLine(); } static int GetLog2_Naive(int v) { int r = 0; while ((v = v >> 1) != 0) { ++r; } return r; } static readonly int[] MultiplyDeBruijnBitPosition2 = new int[32] { 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9 }; static int GetLog2_DeBruijn(int v) { return MultiplyDeBruijnBitPosition2[(uint)(v * 0x077CB531U) >> 27]; } static readonly uint[] b = new uint[] { 0xAAAAAAAA, 0xCCCCCCCC, 0xF0F0F0F0, 0xFF00FF00, 0xFFFF0000}; static int GetLog2_Bithack(int v) { int r = (v & b[0]) == 0 ? 0 : 1; int x = 1 << 4; for (int i = 4; i > 0; i--) // unroll for speed... { if ((v & b[i]) != 0) r |= x; x >>= 1; } return r; } static int GetLog2_Switch(int v) { switch (v) { case 0x00000001: return 0; case 0x00000002: return 1; case 0x00000004: return 2; case 0x00000008: return 3; case 0x00000010: return 4; case 0x00000020: return 5; case 0x00000040: return 6; case 0x00000080: return 7; case 0x00000100: return 8; case 0x00000200: return 9; case 0x00000400: return 10; case 0x00000800: return 11; case 0x00001000: return 12; case 0x00002000: return 13; case 0x00004000: return 14; case 0x00008000: return 15; case 0x00010000: return 16; case 0x00020000: return 17; case 0x00040000: return 18; case 0x00080000: return 19; case 0x00100000: return 20; case 0x00200000: return 21; case 0x00400000: return 22; case 0x00800000: return 23; case 0x01000000: return 24; case 0x02000000: return 25; case 0x04000000: return 26; case 0x08000000: return 27; case 0x10000000: return 28; case 0x20000000: return 29; case 0x40000000: return 30; case int.MinValue: return 31; default: return -1; } }}如果我通过展开循环并使用常量而不是数组查找来优化Bithack代码，则其时间与switch语句方法的时间相同。
static int GetLog2_Bithack(int v){ int r = ((v & 0xAAAAAAAA) != 0) ? 1 : 0; if ((v & 0xFFFF0000) != 0) r |= (1 << 4); if ((v & 0xFF00FF00) != 0) r |= (1 << 3); if ((v & 0xF0F0F0F0) != 0) r |= (1 << 2); if ((v & 0xCCCCCCCC) != 0) r |= (1 << 1); return r;}
