c - 有效地检查 Bitflag 不变性(可能的位旋转)

Question

我有一个 byte 用于存储位标志。我有 8 个标志(每个位一个)，可以分为 4 对 2 个标志，它们是互斥的。我按以下方式排列了位标志:

ABCDEFGH
10011000

当标志 B 也被设置时，不能设置标志 A，反之亦然，因此标志 A 和 B 是互斥的。 标志 A 和 B 可以同时取消设置，只是不能同时设置。相同的规则适用于标志 C 和 D、标志 E 和 F 以及标志 G 和 H。

当前测试用例(C 语言):

#include <stdio.h>

int check(char b) { // used to check invariant
  return ((b&0xC0)==0xC0||(b&0x30)==0x30||(b&0x0C)==0x0C||(b&0x03)==0x03)?0:1;
}
int main() {
  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};

  char truth[256] = {
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  1,1,1,0,1,1,1,0,1,1,1,0,0,0,0,0,
  0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
  0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
  0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
  0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
  0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

  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;
}

上面的check()方法目前通过了所有的测试用例。我想知道是否有更有效的方法来检查这个不变量是否为真，使用一些 bit-twiddling hack。我可能必须每秒多次检查这个不变量，所以效率很重要。如果新的安排允许一些原始安排没有的小技巧，我愿意改变标志的安排。

排列 EX:ABCDEFGH --> AHBGCFDE

Answer 1

我还没有分析过这个，所以不能保证，但你可以试试:

int check(char b)
{
    return ! ( (b << 1) & b & 0xaa );
}

如果您接受非零(不仅仅是 1)为失败，零为通过，您可以取消反转 !。

或者，您可以只使用已经生成的查找表:

int check(char b)
{
    return truth[(unsigned char)b];
}

无论您尝试什么，请介绍一下!

(哦，我建议使用 unsigned chars 而不是 signed chars 来存储像这样的位域，因为这种行为对于位移和 bool 值之类的东西定义得更好。大多数编译器可能会做你期望的，但安全总比后悔好)。