理解strlen实现中的代码

Question

关于glibc strlen中string.hin 的实现我有两个问题.

1. 该实现使用带有"洞"的幻数.我无法理解这是如何工作的.有人可以帮我理解这个片段:

   size_t
   strlen (const char *str)
   {
      const char *char_ptr;
      const unsigned long int *longword_ptr;
      unsigned long int longword, himagic, lomagic;
   
      /* Handle the first few characters by reading one character at a time.
         Do this until CHAR_PTR is aligned on a longword boundary.  */
      for (char_ptr = str; ((unsigned long int) char_ptr
                & (sizeof (longword) - 1)) != 0;
           ++char_ptr)
        if (*char_ptr == '\0')
          return char_ptr - str;
   
      /* All these elucidatory comments refer to 4-byte longwords,
         but the theory applies equally well to 8-byte longwords.  */
   
      longword_ptr = (unsigned long int *) char_ptr;
   
      /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
         the "holes."  Note that there is a hole just to the left of
         each byte, with an extra at the end:
   
         bits:  01111110 11111110 11111110 11111111
         bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
   
         The 1-bits make sure that carries propagate to the next 0-bit.
         The 0-bits provide holes for carries to fall into.  */
   
       himagic = 0x80808080L;
          lomagic = 0x01010101L;
          if (sizeof (longword) > 4)
          {
              /* 64-bit version of the magic.  */
              /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
              himagic = ((himagic << 16) << 16) | himagic;
                lomagic = ((lomagic << 16) << 16) | lomagic;
            }
          if (sizeof (longword) > 8)
            abort ();
   
          /* Instead of the traditional loop which tests each character,
             we will test a longword at a time.  The tricky part is testing
             if *any of the four* bytes in the longword in question are zero.  */
          for (;;)
            {
              longword = *longword_ptr++;
   
              if (((longword - lomagic) & ~longword & himagic) != 0)
            {
              /* Which of the bytes was the zero?  If none of them were, it was
                 a misfire; continue the search.  */
   
              const char *cp = (const char *) (longword_ptr - 1);
   
              if (cp[0] == 0)
                return cp - str;
              if (cp[1] == 0)
                return cp - str + 1;
              if (cp[2] == 0)
                return cp - str + 2;
              if (cp[3] == 0)
                return cp - str + 3;
              if (sizeof (longword) > 4)
                {
                  if (cp[4] == 0)
                return cp - str + 4;
                  if (cp[5] == 0)
                return cp - str + 5;
                  if (cp[6] == 0)
                return cp - str + 6;
        if (cp[7] == 0)
         return cp - str + 7;
   }}}
   

   用于的神奇数字是多少？

2. 为什么不简单地将指针递增到NULL字符并返回计数？这种方法更快吗？为什么会这样？

Answer 1

这用于一次查看4个字节(32位)或甚至8个(64位),以检查其中一个是否为零(字符串结束),而不是单独检查每个字节.

以下是检查空字节的一个示例:

unsigned int v; // 32-bit word to check if any 8-bit byte in it is 0
bool hasZeroByte = ~((((v & 0x7F7F7F7F) + 0x7F7F7F7F) | v) | 0x7F7F7F7F);

对于更多人来说,看看Bit Twiddling Hacks.

这里使用的那个(32位示例):

还有一种更快的方法 - 使用hasless(v,1),定义如下; 它适用于4个操作,不需要后续验证.它简化为

#define haszero(v) (((v) - 0x01010101UL) & ~(v) & 0x80808080UL)

子表达式(v - 0x01010101UL),只要v中的相应字节为零或大于0x80,就会在任何字节中设置为高位设置.子表达式~v&0x80808080UL评估为以字节为单位设置的高位,其中v的字节没有设置其高位(因此字节小于0x80).最后,通过对这两个子表达式进行AND运算,结果是高位设置,其中v中的字节为零,因为由于第一个子表达式中大于0x80的值而设置的高位被第二个子表达式屏蔽掉.

一次查看一个字节的成本至少与查看完整的整数值(寄存器宽)一样多.在该算法中,检查完整的整数以查看它们是否包含零.如果没有,则使用很少的指令,并且可以跳转到下一个完整的整数.如果内部有一个零字节,则进一步检查以查看它的确切位置.