Question

我有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也被置位时，不能设置标志A，反之亦然，因此标志A和B是互斥的。 旗帜A＆amp; B既可以不设置，也可以不设置。标志C＆amp; D，旗帜E＆amp; F和旗帜G＆amp;小时。

当前测试用例（在C中）：

#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。

排列EX：ABCDEFGH - ＆gt; AHBGCFDE

Answer 1

我没有说过这个，所以没有承诺，但你可以尝试：

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，请详细说明！

（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，但比抱歉更安全。

Answer 2

您基本上有四个三态值：设置选项1，或设置选项2，或两者都设置。您可能需要考虑使用结构中的位字段来显式获取每个两位的nsigned整数来跟踪这些三态：

struct Flags {
    unsigned opt1: 2;
    unsigned opt2: 2;
    unsigned opt3: 2;
    unsigned opt4: 2;
};
const unsigned NEITHER = 0;
const unsigned FIRSTOPT = 1;
const unsigned SECONDOPT = 2;
const unsigned INVALIDOPT = 3;

通过此设置，您的问题归结为检查所有四个字段都不等于INVALIDOPT。因此，您的检查功能可以

bool IsValid(struct Flags flags) {
    return flags.opt1 != INVALIDOPT &&
           flags.opt2 != INVALIDOPT &&
           flags.opt3 != INVALIDOPT &&
           flags.opt4 != INVALIDOPT;
}

这也意味着您可以使用命名常量来执行常规变量读取和写入操作，而不是使用twiddling位来设置选择哪个选项。它应该更容易，更安全。

希望这有帮助！

Answer 3

首先要阻止它：

#define NONE_SET 0 // 00
#define FIRST_SET 1 // 10
#define SECOND_SET 2 // 01

int magic(int AB, int CD, int EF, int GH) {
  return (((AB*3+CD)*3+EF)*3+GH)*3
}

更好的是，使用enum代替#define s

示例调用（例如，对于ABCDEFGH = 10000100）

int magic_number = magic(FIRST_SET, NONE_SET, SECOND_SET, NONE_SET);

没有一点麻烦，只是一些乘法。（这可以自动化）

如果您只想检查一个字节中的位对是否都未设置：

uint8_t in; // the input byte
uint8_t out = in & (in >> 1) & 0x55; 
// if out != 0, the invariant is violated

注意：0x55是二进制的01010101（即我们屏蔽了BC之类的相邻对，只留下像AB和CD那样的那些）

有效地检查Bitflag不变量（可能的位旋转）

3 个答案: