glibc 中的 strlen 高效实现

 

#include <string.h>
#include <stdlib.h>

#undef strlen

/* 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(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;
            }
        }
    }
}

/* 说明

(1) 一次判断一个字符直到内存对齐，如果在内存对齐之前就遇到'\0'则直接return，否则到(2)；
(2) 一次读入并判断一个DWORD，如果此DWORD中没有为0的字节，则继续下一个DWORD，否则到(3)；
(3) 到这里则说明DWORD中至少有一个字节为0，剩下的就是找出第一个为0的字节的位置然后return。

*/