(CFLAGS-tst-align.c): Add -mpreferred-stack-boundary=4.

1 files changed, 0 insertions, 153 deletions

diff --git a/string/strlen.c b/string/strlen.c
deleted file mode 100644
index 9bc9db68f7..0000000000
--- a/string/strlen.c
+++ /dev/null
@@ -1,153 +0,0 @@
-/* Copyright (C) 1991, 1993, 1997, 2000, 2003 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, write to the Free
-   Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
-   02111-1307 USA.  */
-
-#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 (str)
-     const char *str;
-{
-  const char *char_ptr;
-  const unsigned long int *longword_ptr;
-  unsigned long int longword, magic_bits, 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.  */
-  magic_bits = 0x7efefeffL;
-  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.  */
-      magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
-      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 (;;)
-    {
-      /* 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 (
-#if 0
-	  /* Add MAGIC_BITS to LONGWORD.  */
-	  (((longword + magic_bits)
-
-	    /* Set those bits that were unchanged by the addition.  */
-	    ^ ~longword)
-
-	   /* Look at only the hole bits.  If any of the hole bits
-	      are unchanged, most likely one of the bytes was a
-	      zero.  */
-	   & ~magic_bits)
-#else
-	  ((longword - lomagic) & himagic)
-#endif
-	  != 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)