--- /dev/null
+/* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006,
+ 2008 Free Software Foundation, Inc.
+
+ Based on strlen implementation 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 in glibc by Roland McGrath (roland@ai.mit.edu).
+ Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).
+
+This program is free software: you can redistribute it and/or modify it
+under the terms of the GNU General Public License as published by the
+Free Software Foundation; either version 3 of the License, or any
+later version.
+
+This program 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 General Public License for more details.
+
+You should have received a copy of the GNU General Public License
+along with this program. If not, see <http://www.gnu.org/licenses/>. */
+
+#include <config.h>
+
+#include "memchr2.h"
+
+#include <limits.h>
+#include <stdint.h>
+#include <string.h>
+
+/* Return the first address of either C1 or C2 (treated as unsigned
+ char) that occurs within N bytes of the memory region S. If
+ neither byte appears, return NULL. */
+void *
+memchr2 (void const *s, int c1_in, int c2_in, size_t n)
+{
+ const unsigned char *char_ptr;
+ const uintmax_t *longword_ptr;
+ uintmax_t longword1;
+ uintmax_t longword2;
+ uintmax_t magic_bits;
+ uintmax_t charmask1;
+ uintmax_t charmask2;
+ unsigned char c1;
+ unsigned char c2;
+ int i;
+
+ c1 = (unsigned char) c1_in;
+ c2 = (unsigned char) c2_in;
+
+ if (c1 == c2)
+ return memchr (s, c1, n);
+
+ /* 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 = (const unsigned char *) s;
+ n > 0 && (size_t) char_ptr % sizeof longword1 != 0;
+ --n, ++char_ptr)
+ if (*char_ptr == c1 || *char_ptr == c2)
+ return (void *) char_ptr;
+
+ /* All these elucidatory comments refer to 4-byte longwords,
+ but the theory applies equally well to any size longwords. */
+
+ longword_ptr = (const uintmax_t *) 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. */
+
+ /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
+ Set CHARMASK to be a longword, each of whose bytes is C. */
+
+ magic_bits = 0xfefefefe;
+ charmask1 = c1 | (c1 << 8);
+ charmask2 = c2 | (c2 << 8);
+ charmask1 |= charmask2 << 16;
+ charmask1 |= charmask2 << 16;
+#if 0xffffffffU < UINTMAX_MAX
+ magic_bits |= magic_bits << 32;
+ charmask1 |= charmask1 << 32;
+ charmask2 |= charmask2 << 32;
+ if (8 < sizeof longword1)
+ for (i = 64; i < sizeof longword1 * 8; i *= 2)
+ {
+ magic_bits |= magic_bits << i;
+ charmask1 |= charmask1 << i;
+ charmask2 |= charmask2 << i;
+ }
+#endif
+ magic_bits = (UINTMAX_MAX >> 1) & (magic_bits | 1);
+
+ /* 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 longword1)
+ {
+ /* 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. */
+
+ longword1 = *longword_ptr ^ charmask1;
+ longword2 = *longword_ptr++ ^ charmask2;
+
+ /* Add MAGIC_BITS to LONGWORD. */
+ if ((((longword1 + magic_bits)
+
+ /* Set those bits that were unchanged by the addition. */
+ ^ ~longword1)
+
+ /* 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
+ || (((longword2 + magic_bits) ^ ~longword2) & ~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] == c1 || cp[0] == c2)
+ return (void *) cp;
+ if (cp[1] == c1 || cp[1] == c2)
+ return (void *) &cp[1];
+ if (cp[2] == c1 || cp[2] == c2)
+ return (void *) &cp[2];
+ if (cp[3] == c1 || cp[3] == c2)
+ return (void *) &cp[3];
+ if (4 < sizeof longword1 && (cp[4] == c1 || cp[4] == c2))
+ return (void *) &cp[4];
+ if (5 < sizeof longword1 && (cp[5] == c1 || cp[5] == c2))
+ return (void *) &cp[5];
+ if (6 < sizeof longword1 && (cp[6] == c1 || cp[6] == c2))
+ return (void *) &cp[6];
+ if (7 < sizeof longword1 && (cp[7] == c1 || cp[7] == c2))
+ return (void *) &cp[7];
+ if (8 < sizeof longword1)
+ for (i = 8; i < sizeof longword1; i++)
+ if (cp[i] == c1 || cp[i] == c2)
+ return (void *) &cp[i];
+ }
+
+ n -= sizeof longword1;
+ }
+
+ char_ptr = (const unsigned char *) longword_ptr;
+
+ while (n-- > 0)
+ {
+ if (*char_ptr == c1 || *char_ptr == c2)
+ return (void *) char_ptr;
+ ++char_ptr;
+ }
+
+ return 0;
+}
--- /dev/null
+/*
+ * Copyright (C) 2008 Free Software Foundation
+ * Written by Eric Blake
+ *
+ * This program is free software: you can redistribute it and/or modify
+ * it under the terms of the GNU General Public License as published by
+ * the Free Software Foundation; either version 3 of the License, or
+ * (at your option) any later version.
+ *
+ * This program 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 General Public License for more details.
+ *
+ * You should have received a copy of the GNU General Public License
+ * along with this program. If not, see <http://www.gnu.org/licenses/>. */
+
+#include <config.h>
+
+#include "memchr2.h"
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+#define ASSERT(expr) \
+ do \
+ { \
+ if (!(expr)) \
+ { \
+ fprintf (stderr, "%s:%d: assertion failed\n", __FILE__, __LINE__); \
+ abort (); \
+ } \
+ } \
+ while (0)
+
+int
+main ()
+{
+ size_t n = 0x100000;
+ char *input = malloc (n);
+ ASSERT (input);
+
+ input[0] = 'a';
+ input[1] = 'b';
+ memset (input + 2, 'c', 1024);
+ memset (input + 1026, 'd', n - 1028);
+ input[n - 2] = 'e';
+ input[n - 1] = 'a';
+
+ ASSERT (memchr2 (input, 'a', 'b', n) == input);
+ ASSERT (memchr2 (input, 'b', 'a', n) == input);
+
+ ASSERT (memchr2 (input, 'a', 'b', 0) == NULL);
+ ASSERT (memchr2 (NULL, 'a', 'b', 0) == NULL);
+
+ ASSERT (memchr2 (input, 'b', 'd', n) == input + 1);
+ ASSERT (memchr2 (input + 2, 'b', 'd', n - 2) == input + 1026);
+
+ ASSERT (memchr2 (input, 'd', 'e', n) == input + 1026);
+ ASSERT (memchr2 (input, 'e', 'd', n) == input + 1026);
+
+ ASSERT (memchr2 (input + 1, 'a', 'e', n - 1) == input + n - 2);
+ ASSERT (memchr2 (input + 1, 'e', 'a', n - 1) == input + n - 2);
+
+ ASSERT (memchr2 (input, 'f', 'g', n) == NULL);
+ ASSERT (memchr2 (input, 'f', '\0', n) == NULL);
+
+ ASSERT (memchr2 (input, 'a', 'a', n) == input);
+ ASSERT (memchr2 (input + 1, 'a', 'a', n - 1) == input + n - 1);
+ ASSERT (memchr2 (input, 'f', 'f', n) == NULL);
+
+ /* Check that a very long haystack is handled quickly if one of the
+ two bytes is found near the beginning. */
+ {
+ size_t repeat = 10000;
+ for (; repeat > 0; repeat--)
+ {
+ ASSERT (memchr2 (input, 'c', 'e', n) == input + 2);
+ ASSERT (memchr2 (input, 'e', 'c', n) == input + 2);
+ ASSERT (memchr2 (input, 'c', '\0', n) == input + 2);
+ ASSERT (memchr2 (input, '\0', 'c', n) == input + 2);
+ }
+ }
+
+ free (input);
+
+ return 0;
+}
projects / pspp / commitdiff