286 lines
6.7 KiB
C
286 lines
6.7 KiB
C
/* Copyright (C) 1991 Free Software Foundation, Inc.
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Contributed by Torbjorn Granlund (tege@sics.se).
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The GNU C Library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Library General Public License as
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published by the Free Software Foundation; either version 2 of the
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License, or (at your option) any later version.
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The GNU C Library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Library General Public License for more details.
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You should have received a copy of the GNU Library General Public
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License along with the GNU C Library; see the file COPYING.LIB. If
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not, write to the Free Software Foundation, Inc., 675 Mass Ave,
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Cambridge, MA 02139, USA. */
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#include <ansidecl.h>
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#include <string.h>
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#include <memcopy.h>
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/* BE VERY CAREFUL IF YOU CHANGE THIS CODE! */
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/* The strategy of this memcmp is:
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1. Compare bytes until one of the block pointers is aligned.
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2. Compare using memcmp_common_alignment or
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memcmp_not_common_alignment, regarding the alignment of the other
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block after the initial byte operations. The maximum number of
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full words (of type op_t) are compared in this way.
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3. Compare the few remaining bytes. */
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/* memcmp_common_alignment -- Compare blocks at SRCP1 and SRCP2 with LEN `op_t'
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objects (not LEN bytes!). Both SRCP1 and SRCP2 should be aligned for
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memory operations on `op_t's. */
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#ifdef __GNUC__
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__inline
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#endif
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static int
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DEFUN(memcmp_common_alignment, (srcp1, srcp2, len),
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long int srcp1 AND long int srcp2 AND size_t len)
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{
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op_t a0, a1;
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op_t b0, b1;
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op_t res;
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switch (len % 4)
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{
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case 2:
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a0 = ((op_t *) srcp1)[0];
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b0 = ((op_t *) srcp2)[0];
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srcp1 -= 2 * OPSIZ;
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srcp2 -= 2 * OPSIZ;
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len += 2;
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goto do1;
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case 3:
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a1 = ((op_t *) srcp1)[0];
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b1 = ((op_t *) srcp2)[0];
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srcp1 -= OPSIZ;
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srcp2 -= OPSIZ;
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len += 1;
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goto do2;
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case 0:
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if (OP_T_THRES <= 3 * OPSIZ && len == 0)
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return 0;
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a0 = ((op_t *) srcp1)[0];
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b0 = ((op_t *) srcp2)[0];
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goto do3;
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case 1:
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a1 = ((op_t *) srcp1)[0];
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b1 = ((op_t *) srcp2)[0];
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srcp1 += OPSIZ;
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srcp2 += OPSIZ;
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len -= 1;
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if (OP_T_THRES <= 3 * OPSIZ && len == 0)
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goto do0;
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/* Fall through. */
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}
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do
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{
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a0 = ((op_t *) srcp1)[0];
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b0 = ((op_t *) srcp2)[0];
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res = a1 - b1;
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if (res != 0)
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return res;
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do3:
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a1 = ((op_t *) srcp1)[1];
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b1 = ((op_t *) srcp2)[1];
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res = a0 - b0;
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if (res != 0)
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return res;
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do2:
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a0 = ((op_t *) srcp1)[2];
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b0 = ((op_t *) srcp2)[2];
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res = a1 - b1;
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if (res != 0)
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return res;
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do1:
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a1 = ((op_t *) srcp1)[3];
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b1 = ((op_t *) srcp2)[3];
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res = a0 - b0;
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if (res != 0)
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return res;
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srcp1 += 4 * OPSIZ;
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srcp2 += 4 * OPSIZ;
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len -= 4;
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}
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while (len != 0);
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/* This is the right position for do0. Please don't move
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it into the loop. */
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do0:
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return a1 - b1;
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}
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/* SRCP2 should be aligned for memory operations on `op_t',
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but SRCP1 *should be unaligned*. */
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#ifdef __GNUC__
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__inline
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#endif
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static int
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DEFUN(memcmp_not_common_alignment, (srcp1, srcp2, len),
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long int srcp1 AND long int srcp2 AND size_t len)
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{
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op_t a0, a1, a2, a3;
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op_t b0, b1, b2, b3;
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op_t res;
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op_t x;
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int shl, shr;
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/* Calculate how to shift a word read at the memory operation
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aligned srcp1 to make it aligned for comparison. */
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shl = 8 * (srcp1 % OPSIZ);
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shr = 8 * OPSIZ - shl;
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/* Make SRCP1 aligned by rounding it down to the beginning of the `op_t'
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it points in the middle of. */
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srcp1 &= -OPSIZ;
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switch (len % 4)
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{
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case 2:
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a1 = ((op_t *) srcp1)[0];
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a2 = ((op_t *) srcp1)[1];
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b2 = ((op_t *) srcp2)[0];
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srcp1 -= 1 * OPSIZ;
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srcp2 -= 2 * OPSIZ;
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len += 2;
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goto do1;
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case 3:
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a0 = ((op_t *) srcp1)[0];
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a1 = ((op_t *) srcp1)[1];
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b1 = ((op_t *) srcp2)[0];
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srcp2 -= 1 * OPSIZ;
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len += 1;
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goto do2;
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case 0:
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if (OP_T_THRES <= 3 * OPSIZ && len == 0)
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return 0;
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a3 = ((op_t *) srcp1)[0];
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a0 = ((op_t *) srcp1)[1];
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b0 = ((op_t *) srcp2)[0];
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srcp1 += 1 * OPSIZ;
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goto do3;
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case 1:
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a2 = ((op_t *) srcp1)[0];
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a3 = ((op_t *) srcp1)[1];
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b3 = ((op_t *) srcp2)[0];
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srcp1 += 2 * OPSIZ;
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srcp2 += 1 * OPSIZ;
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len -= 1;
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if (OP_T_THRES <= 3 * OPSIZ && len == 0)
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goto do0;
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/* Fall through. */
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}
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do
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{
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a0 = ((op_t *) srcp1)[0];
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b0 = ((op_t *) srcp2)[0];
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x = MERGE(a2, shl, a3, shr);
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res = x - b3;
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if (res != 0)
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return res;
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do3:
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a1 = ((op_t *) srcp1)[1];
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b1 = ((op_t *) srcp2)[1];
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x = MERGE(a3, shl, a0, shr);
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res = x - b0;
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if (res != 0)
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return res;
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do2:
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a2 = ((op_t *) srcp1)[2];
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b2 = ((op_t *) srcp2)[2];
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x = MERGE(a0, shl, a1, shr);
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res = x - b1;
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if (res != 0)
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return res;
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do1:
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a3 = ((op_t *) srcp1)[3];
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b3 = ((op_t *) srcp2)[3];
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x = MERGE(a1, shl, a2, shr);
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res = x - b2;
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if (res != 0)
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return res;
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srcp1 += 4 * OPSIZ;
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srcp2 += 4 * OPSIZ;
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len -= 4;
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}
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while (len != 0);
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/* This is the right position for do0. Please don't move
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it into the loop. */
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do0:
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x = MERGE(a2, shl, a3, shr);
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return x - b3;
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}
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int
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DEFUN(memcmp, (s1, s2, n),
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CONST PTR s1 AND CONST PTR s2 AND size_t len)
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{
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op_t a0;
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op_t b0;
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long int srcp1 = (long int) s1;
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long int srcp2 = (long int) s2;
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op_t res;
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if (len >= OP_T_THRES)
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{
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/* There are at least some bytes to compare. No need to test
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for LEN == 0 in this alignment loop. */
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while (srcp2 % OPSIZ != 0)
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{
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a0 = ((byte *) srcp1)[0];
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b0 = ((byte *) srcp2)[0];
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srcp1 += 1;
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srcp2 += 1;
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res = a0 - b0;
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if (res != 0)
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return res;
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len -= 1;
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}
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/* SRCP2 is now aligned for memory operations on `op_t'.
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SRCP1 alignment determines if we can do a simple,
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aligned compare or need to shuffle bits. */
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if (srcp1 % OPSIZ == 0)
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res = memcmp_common_alignment (srcp1, srcp2, len / OPSIZ);
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else
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res = memcmp_not_common_alignment (srcp1, srcp2, len / OPSIZ);
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if (res != 0)
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return res;
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/* Number of bytes remaining in the interval [0..OPSIZ-1]. */
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srcp1 += len & -OPSIZ;
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srcp2 += len & -OPSIZ;
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len %= OPSIZ;
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}
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/* There are just a few bytes to compare. Use byte memory operations. */
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while (len != 0)
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{
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a0 = ((byte *) srcp1)[0];
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b0 = ((byte *) srcp2)[0];
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srcp1 += 1;
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srcp2 += 1;
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res = a0 - b0;
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if (res != 0)
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return res;
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len -= 1;
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}
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return 0;
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}
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