optimization - 确定字节中最高有效位的位置

Question

我有一个 byte 用于存储位标志。我需要计算字节中最高有效位的位置。

示例字节: 00101101 => 6 是最高位设置位的位置

紧凑的十六进制映射:

[0x00]      => 0x00
[0x01]      => 0x01
[0x02,0x03] => 0x02
[0x04,0x07] => 0x03
[0x08,0x0F] => 0x04
[0x10,0x1F] => 0x05
[0x20,0x3F] => 0x06
[0x40,0x7F] => 0x07
[0x80,0xFF] => 0x08

C 中的测试用例:

#include <stdio.h>

unsigned char check(unsigned char b) {
  unsigned char c = 0x08;
  unsigned char m = 0x80;
  do {
    if(m&b) { return  c; }
    else    { c -= 0x01; }
  } while(m>>=1);
  return 0; //never reached
}
int main() {
  unsigned char input[256] = {
    0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,
    0x10,0x11,0x12,0x13,0x14,0x15,0x16,0x17,0x18,0x19,0x1a,0x1b,0x1c,0x1d,0x1e,0x1f,
    0x20,0x21,0x22,0x23,0x24,0x25,0x26,0x27,0x28,0x29,0x2a,0x2b,0x2c,0x2d,0x2e,0x2f,
    0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x3a,0x3b,0x3c,0x3d,0x3e,0x3f,
    0x40,0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x4f,
    0x50,0x51,0x52,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x5b,0x5c,0x5d,0x5e,0x5f,
    0x60,0x61,0x62,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x6a,0x6b,0x6c,0x6d,0x6e,0x6f,
    0x70,0x71,0x72,0x73,0x74,0x75,0x76,0x77,0x78,0x79,0x7a,0x7b,0x7c,0x7d,0x7e,0x7f,
    0x80,0x81,0x82,0x83,0x84,0x85,0x86,0x87,0x88,0x89,0x8a,0x8b,0x8c,0x8d,0x8e,0x8f,
    0x90,0x91,0x92,0x93,0x94,0x95,0x96,0x97,0x98,0x99,0x9a,0x9b,0x9c,0x9d,0x9e,0x9f,
    0xa0,0xa1,0xa2,0xa3,0xa4,0xa5,0xa6,0xa7,0xa8,0xa9,0xaa,0xab,0xac,0xad,0xae,0xaf,
    0xb0,0xb1,0xb2,0xb3,0xb4,0xb5,0xb6,0xb7,0xb8,0xb9,0xba,0xbb,0xbc,0xbd,0xbe,0xbf,
    0xc0,0xc1,0xc2,0xc3,0xc4,0xc5,0xc6,0xc7,0xc8,0xc9,0xca,0xcb,0xcc,0xcd,0xce,0xcf,
    0xd0,0xd1,0xd2,0xd3,0xd4,0xd5,0xd6,0xd7,0xd8,0xd9,0xda,0xdb,0xdc,0xdd,0xde,0xdf,
    0xe0,0xe1,0xe2,0xe3,0xe4,0xe5,0xe6,0xe7,0xe8,0xe9,0xea,0xeb,0xec,0xed,0xee,0xef,
    0xf0,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf8,0xf9,0xfa,0xfb,0xfc,0xfd,0xfe,0xff };

  unsigned char truth[256] = {
    0x00,0x01,0x02,0x02,0x03,0x03,0x03,0x03,0x04,0x04,0x04,0x04,0x04,0x04,0x04,0x04, 
    0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05, 
    0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06, 
    0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08};

  int i,r;
  int f = 0;
  for(i=0; i<256; ++i) {
    r=check(input[i]);
    if(r !=(truth[i])) {
      printf("failed %d : 0x%x : %d\n",i,0x000000FF & ((int)input[i]),r);
      f += 1;
    }
  }
  if(!f) { printf("passed all\n");  }
  else   { printf("failed %d\n",f); }
  return 0;
}

我想简化我的 check() 函数，使其不涉及循环(最好是分支)。是否有一些小技巧或散列查找表解决方案来计算一个字节中最高有效位的位置？

Answer 1

您的问题是关于计算值的 log2 的有效方法。因为您似乎想要一个不限于 C 语言的解决方案，所以我有点懒惰并调整了一些我的 C# 代码。

您想计算 log2(x) + 1 并且对于 x = 0(其中 log2 未定义)您将结果定义为0(例如，您创建了一个特殊情况，其中 log2(0) = -1)。

static readonly Byte[] multiplyDeBruijnBitPosition = new Byte[] {
  7, 2, 3, 4,
  6, 1, 5, 0
};

public static Byte Log2Plus1(Byte value) {
  if (value == 0)
    return 0;

  var roundedValue = value;
  roundedValue |= (Byte) (roundedValue >> 1);
  roundedValue |= (Byte) (roundedValue >> 2);
  roundedValue |= (Byte) (roundedValue >> 4);
  var log2 = multiplyDeBruijnBitPosition[((Byte) (roundedValue*0xE3)) >> 5];
  return (Byte) (log2 + 1);
}

这个小技巧取自 Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup您可以在其中查看 32 位值的等效 C 源代码。此代码已经过调整以处理 8 位值。

但是，您可以使用一个非常有效的内置函数来使用一个操作来为您提供结果(在许多 CPU 上使用一条指令，如 Bit Scan Reverse)。问题的答案 Bit twiddling: which bit is set?有一些关于这个的信息。答案中的一句话提供了解决此问题的低级别支持的一个可能原因:

Things like this are the core of many O(1) algorithms such as kernel schedulers which need to find the first non-empty queue signified by an array of bits.