glibc-2.19 之 strlen 实现

前几天遇到一个有意思的问题，实现strlen 不考虑线程安全:

下面是我的实现:

1 size_t strlen(const char* s)
2 {
3     const char* p = s;
4     while (*p++);
5     return p-1-s;
6 }

Glibc 2.19 的实现:
针对此实现，函数头部分没太明白， size_t strlen (str) const char *str; 详细情况参见Glibc 2.19, 下面的实现还是比较经典的兼顾性能和体系机构。

 1 /* Return the length of the null-terminated string STR.  Scan for
 2    the null terminator quickly by testing four bytes at a time.  */
 3 size_t
 4 strlen (str)
 5      const char *str;
 6 {
 7   const char *char_ptr;
 8   const unsigned long int *longword_ptr;
 9   unsigned long int longword, himagic, lomagic;
10 
11   /* Handle the first few characters by reading one character at a time.
12      Do this until CHAR_PTR is aligned on a longword boundary.  */
13   for (char_ptr = str; ((unsigned long int) char_ptr
14             & (sizeof (longword) - 1)) != 0;
15        ++char_ptr)
16     if (*char_ptr == '\0')
17       return char_ptr - str;
18 
19   /* All these elucidatory comments refer to 4-byte longwords,
20      but the theory applies equally well to 8-byte longwords.  */
21 
22   longword_ptr = (unsigned long int *) char_ptr;
23 
24   /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
25      the "holes."  Note that there is a hole just to the left of
26      each byte, with an extra at the end:
27 
28      bits:  01111110 11111110 11111110 11111111
29      bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
30 
31      The 1-bits make sure that carries propagate to the next 0-bit.
32      The 0-bits provide holes for carries to fall into.  */
33   himagic = 0x80808080L;
34   lomagic = 0x01010101L;
35   if (sizeof (longword) > 4)
36     {
37       /* 64-bit version of the magic.  */
38       /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
39       himagic = ((himagic << 16) << 16) | himagic;
40       lomagic = ((lomagic << 16) << 16) | lomagic;
41     }
42   if (sizeof (longword) > 8)
43     abort ();
44 
45   /* Instead of the traditional loop which tests each character,
46      we will test a longword at a time.  The tricky part is testing
47      if *any of the four* bytes in the longword in question are zero.  */
48   for (;;)
49     {
50       longword = *longword_ptr++;
51 
52       if (((longword - lomagic) & ~longword & himagic) != 0)
53     {
54       /* Which of the bytes was the zero?  If none of them were, it was
55          a misfire; continue the search.  */
56 
57       const char *cp = (const char *) (longword_ptr - 1);
58 
59       if (cp[0] == 0)
60         return cp - str;
61       if (cp[1] == 0)
62         return cp - str + 1;
63       if (cp[2] == 0)
64         return cp - str + 2;
65       if (cp[3] == 0)
66         return cp - str + 3;
67       if (sizeof (longword) > 4)
68         {
69           if (cp[4] == 0)
70         return cp - str + 4;
71           if (cp[5] == 0)
72         return cp - str + 5;
73           if (cp[6] == 0)
74         return cp - str + 6;
75           if (cp[7] == 0)
76         return cp - str + 7;
77         }
78     }
79     }
80 }

 

posted @ 2014-07-27 13:03  Yuangui  阅读(289)  评论(0)  编辑  收藏  举报