有效地检查Bitflag不变量(可能的位旋转)

Question

我有一个byte我用来存储位标志.我有8个标志(每个位一个),可以分成4对2个标志,它们是互斥的.我已按以下方式安排了位标志:

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都可以取消设置,而不是两者都设置.标志C&D,标志E&F和标志G&H也是如此.

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;
}

上述check()方法目前通过了所有测试用例.我想知道是否有一种更有效的方法来检查这个不变量是否正确使用一些比特笨拙的黑客.我可能需要每秒多次检查这个不变量,因此效率很重要.如果新的安排允许原始安排没有进行一些麻烦的黑客攻击,我愿意置换旗帜的安排.

置换EX:ABCDEFGH- >AHBGCFDE

Answer 1

我没有描述过这个,所以没有承诺,但你可以试试:

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

!如果你接受非零(不仅仅是1)作为失败,你可以取消反转,并将零作为通过.

或者,您可以使用已生成的查找表:

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

无论您尝试什么,请介绍!

(哦,我建议使用无符号字符而不是有符号字符来存储像这样的位字段,因为对于像位移和布尔值这样的行为更好地定义.大多数编译器可能会按照你的期望做,但更安全不好意思).