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authorRoland McGrath <roland@gnu.org>1995-02-18 01:27:10 +0000
committerRoland McGrath <roland@gnu.org>1995-02-18 01:27:10 +0000
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+/* Copyright (C) 1991, 1993 Free Software Foundation, Inc.
+ Based on strlen implemention by Torbjorn Granlund (tege@sics.se),
+ with help from Dan Sahlin (dan@sics.se) and
+ commentary by Jim Blandy (jimb@ai.mit.edu);
+ adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
+ and implemented by Roland McGrath (roland@ai.mit.edu).
+
+The GNU C Library is free software; you can redistribute it and/or
+modify it under the terms of the GNU Library General Public License as
+published by the Free Software Foundation; either version 2 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
+Library General Public License for more details.
+
+You should have received a copy of the GNU Library General Public
+License along with the GNU C Library; see the file COPYING.LIB. If
+not, write to the Free Software Foundation, Inc., 675 Mass Ave,
+Cambridge, MA 02139, USA. */
+
+#include <ansidecl.h>
+#include <string.h>
+
+
+/* Search no more than N bytes of S for C. */
+
+PTR
+DEFUN(memchr, (s, c, n), CONST PTR s AND int c AND size_t n)
+{
+ CONST unsigned char *char_ptr;
+ CONST unsigned long int *longword_ptr;
+ unsigned long int longword, magic_bits, charmask;
+
+ c = (unsigned char) c;
+
+ /* 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 = s; n > 0 && ((unsigned long int) char_ptr
+ & (sizeof (longword) - 1)) != 0;
+ --n, ++char_ptr)
+ if (*char_ptr == c)
+ return (PTR) char_ptr;
+
+ /* 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. */
+ switch (sizeof (longword))
+ {
+ case 4: magic_bits = 0x7efefeffL; break;
+ case 8: magic_bits = (0x7efefefeL << 32) | 0xfefefeffL; break;
+ default:
+ abort ();
+ }
+
+ /* Set up a longword, each of whose bytes is C. */
+ charmask = c | (c << 8);
+ charmask |= charmask << 16;
+ if (sizeof (longword) > 4)
+ charmask |= charmask << 32;
+ 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 (n >= sizeof (longword))
+ {
+ /* 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.
+
+ 3) But wait! Aren't we looking for C, not zero?
+ Good point. So what we do is XOR LONGWORD with a longword,
+ each of whose bytes is C. This turns each byte that is C
+ into a zero. */
+
+ longword = *longword_ptr++ ^ charmask;
+
+ /* Add MAGIC_BITS to LONGWORD. */
+ if ((((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) != 0)
+ {
+ /* Which of the bytes was C? If none of them were, it was
+ a misfire; continue the search. */
+
+ CONST unsigned char *cp = (CONST unsigned char *) (longword_ptr - 1);
+
+ if (cp[0] == c)
+ return (PTR) cp;
+ if (cp[1] == c)
+ return (PTR) &cp[1];
+ if (cp[2] == c)
+ return (PTR) &cp[2];
+ if (cp[3] == c)
+ return (PTR) &cp[3];
+ if (sizeof (longword) > 4)
+ {
+ if (cp[4] == c)
+ return (PTR) &cp[4];
+ if (cp[5] == c)
+ return (PTR) &cp[5];
+ if (cp[6] == c)
+ return (PTR) &cp[6];
+ if (cp[7] == c)
+ return (PTR) &cp[7];
+ }
+ }
+
+ n -= sizeof (longword);
+ }
+
+ char_ptr = (CONST unsigned char *) longword_ptr;
+
+ while (n-- > 0)
+ {
+ if (*char_ptr == c)
+ return (PTR) char_ptr;
+ else
+ ++char_ptr;
+ }
+
+ return NULL;
+}