/* Copyright (C) 1991-2017 Free Software Foundation, Inc.
   This file is part of the GNU C Library.
   Written by Torbjorn Granlund (tege@sics.se),
   with help from Dan Sahlin (dan@sics.se);
   commentary by Jim Blandy (jimb@ai.mit.edu).

   The GNU C Library is free software; you can redistribute it and/or
   modify it under the terms of the GNU Lesser General Public
   License as published by the Free Software Foundation; either
   version 2.1 of the License, or (at your option) any later version.

   The GNU C Library is distributed in the hope that it will be useful,
   but WITHOUT ANY WARRANTY; without even the implied warranty of
   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
   Lesser General Public License for more details.

   You should have received a copy of the GNU Lesser General Public
   License along with the GNU C Library; if not, see
   <http://www.gnu.org/licenses/>.  */

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

#undef strlen

#ifndef STRLEN
# define STRLEN strlen
#endif

/* 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;
        }
    }
    }
}
libc_hidden_builtin_def (strlen)

上面是glibc-2.25的strlen函數(shù)的的源代碼，我這里呢精簡一下，因為我用的平臺是64位的砖织，所以我就省略了32位的一些判斷，順便加了一些注釋末荐，方便理解

unsigned long long strlen (const char *str){
    for (const char *char_ptr = str; (unsigned long long) char_ptr & 7; ++char_ptr)
        if (*char_ptr == 0)
            return char_ptr - str;
    /*
    這里可能有人不懂侧纯，char_ptr & 7這是在做什么
    解釋一下，這里考慮的一個對齊的甲脏，這個循環(huán)最多只能執(zhí)行7次
    分別是001眶熬，010妹笆，011，100娜氏，101拳缠，110，111贸弥，
    即char_tr的低三位為如上值時
    */
    for (const unsigned long long *longword_ptr = (unsigned long long*) str;;++longword_ptr){
        unsigned long long longword = *longword_ptr;
        if ((longword - 0X0101010101010101) & ~longword & 0X8080808080808080){
            for (unsigned long long i = 0; i < 8; ++i) {
                const char *cp = (const char *)longword_ptr;
                if(cp[i] == 0)
                    return cp - str + i;
            }
            /*
            上面也有一個難點：就是那個條件
            上面的是64位的8字的窟坐，我用8字的char 講一下
            char c;
            ~c & 0X80
            這個的結果只有兩種
            一種是0X80，當且僅當~C的最高為1時就是C的最高位為0時成立
            一種是0
            (c-1)& 0X80的結果也只有兩種
            一種是0X80茂腥，當且僅當(c-1)的最高為1時成立
            一種是0
            要使結果是非0狸涌，即0X80，只要同時滿足兩種條件
            (c-1)& 0X80 & (~c & 0X80)最岗！= 0 即(c-1)& ~c & 0X80同時成立
            即c的最高位是0帕胆，而且(c-1)的最高位是1，當且僅當c等于0時成立
            */
        }
    }
}

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