理解strlen实现中的代码

时间:2016-01-06 21:20:45

标签: c string algorithm pointers gcc

关于glibc中strlenstring.h的实施,我有两个问题。

  1. 该实现使用带有'孔的幻数。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;
    }}}
    

    用于的神奇数字是什么?

  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?

1 个答案:

答案 0 :(得分:14)

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);

如需更多信息,请参阅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),这是定义的   下面;它适用于4个操作,不需要后续操作   验证。它简化为

     

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

     

子表达式(v - 0x01010101UL),计算为设置的高位   任何字节,只要v中的相应字节为零或大于   0x80的。子表达式~v&amp; 0x80808080UL评估为高位设置   以字节为单位,其中v的字节没有设置其高位(所以   字节小于0x80)。最后,通过ANDing这两个子表达式   结果是高位设置,其中v中的字节为零,因为   由于第一个值大于0x80而设置的高位   子表达式被第二个掩盖。

一次查看一个字节的成本至少与查看完整的整数值(寄存器宽度)一样多。在此算法中,检查完整整数以查看它们是否包含零。如果没有,则使用很少的指令,并且可以对下一个完整整数进行跳转。如果内部有一个零字节,则进一步检查以确定它的确切位置。