/* Find the length of STRING, but scan at most MAXLEN characters.

   Copyright (C) 1991-2018 Free Software Foundation, Inc.

   Contributed by Jakub Jelinek <jakub@redhat.com>.

   Based on strlen 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; see the file COPYING.LIB.  If

   not, see <http://www.gnu.org/licenses/>.  */

#include <string.h>

#include <stdlib.h>

/* Find the length of S, but scan at most MAXLEN characters.  If no

   '\0' terminator is found in that many characters, return MAXLEN.  */

#ifdef STRNLEN

# define __strnlen STRNLEN

#endif

size_t

__strnlen (const char *str, size_t maxlen)

{

  const char *char_ptr, *end_ptr = str + maxlen;

  const unsigned long int *longword_ptr;

  unsigned long int longword, himagic, lomagic;

  if (maxlen == 0)

    return 0;

  if (__glibc_unlikely (end_ptr < str))

    end_ptr = (const char *) ~0UL;

  /* 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')

      {

	if (char_ptr > end_ptr)

	  char_ptr = end_ptr;

	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.  */

  while (longword_ptr < (unsigned long int *) end_ptr)

    {

      /* We tentatively exit the loop if adding MAGIC_BITS to

	 LONGWORD fails to change any of the hole bits of LONGWORD.

	 1) Is this safe?  Will it catch all the zero bytes?

	 Suppose there is a byte with all zeros.  Any carry bits

	 propagating from its left will fall into the hole at its

	 least significant bit and stop.  Since there will be no

	 carry from its most significant bit, the LSB of the

	 byte to the left will be unchanged, and the zero will be

	 detected.

	 2) Is this worthwhile?  Will it ignore everything except

	 zero bytes?  Suppose every byte of LONGWORD has a bit set

	 somewhere.  There will be a carry into bit 8.  If bit 8

	 is set, this will carry into bit 16.  If bit 8 is clear,

	 one of bits 9-15 must be set, so there will be a carry

	 into bit 16.  Similarly, there will be a carry into bit

	 24.  If one of bits 24-30 is set, there will be a carry

	 into bit 31, so all of the hole bits will be changed.

	 The one misfire occurs when bits 24-30 are clear and bit

	 31 is set; in this case, the hole at bit 31 is not

	 changed.  If we had access to the processor carry flag,

	 we could close this loophole by putting the fourth hole

	 at bit 32!

	 So it ignores everything except 128's, when they're aligned

	 properly.  */

      longword = *longword_ptr++;

      if ((longword - lomagic) & himagic)

	{

	  /* 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);

	  char_ptr = cp;

	  if (cp[0] == 0)

	    break;

	  char_ptr = cp + 1;

	  if (cp[1] == 0)

	    break;

	  char_ptr = cp + 2;

	  if (cp[2] == 0)

	    break;

	  char_ptr = cp + 3;

	  if (cp[3] == 0)

	    break;

	  if (sizeof (longword) > 4)

	    {

	      char_ptr = cp + 4;

	      if (cp[4] == 0)

		break;

	      char_ptr = cp + 5;

	      if (cp[5] == 0)

		break;

	      char_ptr = cp + 6;

	      if (cp[6] == 0)

		break;

	      char_ptr = cp + 7;

	      if (cp[7] == 0)

		break;

	    }

	}

      char_ptr = end_ptr;

    }

  if (char_ptr > end_ptr)

    char_ptr = end_ptr;

  return char_ptr - str;

}

#ifndef STRNLEN

libc_hidden_def (__strnlen)

weak_alias (__strnlen, strnlen)

#endif

libc_hidden_def (strnlen)