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:

size_t strlen(const char* s)
{
const char* p = s;
while (*p++);
return p-1-s;
}

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。

/* Return the length of the null-terminated string STR.  Scan for
the null terminator quickly by testing four bytes at a time.  */
size_t
strlen (str)
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;
}
}
}
}