glibc strlen 实现方式分析

摘要

本文对 glibc 2.35 源码中的 strlen 函数实现展开研究，分析其使用位运算实现更快长度计算的原理。

函数位置

glibc-2.35/string/strlen.c

原始实现

下面是 glibc strlen.c 的文件内容：

/* Copyright (C) 1991-2022 Free Software Foundation, Inc. This file is part of the GNU C Library. 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 <https://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)

分析

区别于逐一比较每个字符是否为0，glibc的strlen采用每8字节（64位程序）一比较的方式，快速找到为0的字节。之后，对含0的8字节内进行逐字节比较，找到0的位置。

对齐

为提升效率，首先对前几个字符进行逐一比较，直到指针与long型地址对齐。

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;

比较原理

之后，是每8位一比较的原理。
假如某字节为全0，即为 00000000（二进制）。那么，如果我们对它进行减1的操作，它会变为11111111。
注意最高位。如果我们将原字节0进行取反，得到的也是11111111，那么两个结果取与运算，得到的值将会是最高位为1的值。注意，即便紧跟其后的字节在做减法时要借位，该字节只会变成11111110，最高位依旧是1。

---

如果该字节不全为0，则可分两种情况：首位为1和首位为0.
如果该字节首位为0，则其为0abcdef，a-f中至少有1个为1，因此对最低位减1后，首位依然是0.
如果该字节首位为1，且为该字节二进制表示中的唯一一个1，则做减1操作时，首位会变成0。如果不是唯一的1，则首位仍然是1。

---

综合以上，可以看出，如果某字节最低位减1后，最高位为1，且字节取反后最高位也是1，那么该字节一定是0。基于此，我们可以对字节做批量比较。

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;
		}
	}
}