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181 lines
5.0 KiB
ArmAsm
181 lines
5.0 KiB
ArmAsm
/* rawmemchr (str, ch) -- Return pointer to first occurrence of CH in STR.
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For Motorola 68000.
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Copyright (C) 1999, 2002 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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Contributed by Andreas Schwab <schwab@gnu.org>.
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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 Lesser General Public
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License as published by the Free Software Foundation; either
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version 2.1 of the 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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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with the GNU C Library; if not, write to the Free
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Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
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02111-1307 USA. */
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#include <sysdep.h>
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#include "asm-syntax.h"
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TEXT
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ENTRY(__rawmemchr)
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/* Save the callee-saved registers we use. */
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movel R(d2),MEM_PREDEC(sp)
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movel R(d3),MEM_PREDEC(sp)
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/* Get string pointer and character. */
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movel MEM_DISP(sp,12),R(a0)
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moveb MEM_DISP(sp,19),R(d0)
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/* Distribute the character to all bytes of a longword. */
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movel R(d0),R(d1)
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lsll #8,R(d1)
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moveb R(d0),R(d1)
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movel R(d1),R(d0)
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swap R(d0)
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movew R(d1),R(d0)
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/* First search for the character one byte at a time until the
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pointer is aligned to a longword boundary. */
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movel R(a0),R(d1)
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andw #3,R(d1)
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beq L(L1)
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cmpb MEM(a0),R(d0)
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beq L(L9)
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addql #1,R(a0)
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movel R(a0),R(d1)
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andw #3,R(d1)
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beq L(L1)
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cmpb MEM(a0),R(d0)
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beq L(L9)
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addql #1,R(a0)
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movel R(a0),R(d1)
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andw #3,R(d1)
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beq L(L1)
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cmpb MEM(a0),R(d0)
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beq L(L9)
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addql #1,R(a0)
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L(L1:)
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/* Load the magic bits. Unlike the generic implementation we can
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use the carry bit as the fourth hole. */
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movel #0xfefefeff,R(d3)
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/* We exit the loop if adding MAGIC_BITS to LONGWORD fails to
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change any of the hole bits of LONGWORD.
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1) Is this safe? Will it catch all the zero bytes?
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Suppose there is a byte with all zeros. Any carry bits
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propagating from its left will fall into the hole at its
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least significant bit and stop. Since there will be no
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carry from its most significant bit, the LSB of the
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byte to the left will be unchanged, and the zero will be
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detected.
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2) Is this worthwhile? Will it ignore everything except
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zero bytes? Suppose every byte of LONGWORD has a bit set
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somewhere. There will be a carry into bit 8. If bit 8
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is set, this will carry into bit 16. If bit 8 is clear,
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one of bits 9-15 must be set, so there will be a carry
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into bit 16. Similarly, there will be a carry into bit
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24. If one of bits 24-31 is set, there will be a carry
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into bit 32 (=carry flag), so all of the hole bits will
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be changed.
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3) But wait! Aren't we looking for C, not zero?
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Good point. So what we do is XOR LONGWORD with a longword,
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each of whose bytes is C. This turns each byte that is C
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into a zero. */
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L(L2:)
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/* Get the longword in question. */
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movel MEM_POSTINC(a0),R(d1)
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/* XOR with the byte we search for. */
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eorl R(d0),R(d1)
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/* Add the magic value. We get carry bits reported for each byte
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which is not C. */
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movel R(d3),R(d2)
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addl R(d1),R(d2)
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/* Check the fourth carry bit before it is clobbered by the next
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XOR. If it is not set we have a hit. */
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bcc L(L8)
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/* We are only interested in carry bits that change due to the
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previous add, so remove original bits. */
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eorl R(d1),R(d2)
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/* Now test for the other three overflow bits.
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Set all non-carry bits. */
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orl R(d3),R(d2)
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/* Add 1 to get zero if all carry bits were set. */
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addql #1,R(d2)
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/* If we don't get zero then at least one byte of the word equals
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C. */
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bne L(L8)
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/* Get the longword in question. */
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movel MEM_POSTINC(a0),R(d1)
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/* XOR with the byte we search for. */
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eorl R(d0),R(d1)
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/* Add the magic value. We get carry bits reported for each byte
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which is not C. */
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movel R(d3),R(d2)
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addl R(d1),R(d2)
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/* Check the fourth carry bit before it is clobbered by the next
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XOR. If it is not set we have a hit. */
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bcc L(L8)
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/* We are only interested in carry bits that change due to the
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previous add, so remove original bits */
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eorl R(d1),R(d2)
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/* Now test for the other three overflow bits.
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Set all non-carry bits. */
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orl R(d3),R(d2)
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/* Add 1 to get zero if all carry bits were set. */
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addql #1,R(d2)
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/* If we don't get zero then at least one byte of the word equals
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C. */
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beq L(L2)
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L(L8:)
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/* We have a hit. Check to see which byte it was. First
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compensate for the autoincrement in the loop. */
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subql #4,R(a0)
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cmpb MEM(a0),R(d0)
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beq L(L9)
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addql #1,R(a0)
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cmpb MEM(a0),R(d0)
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beq L(L9)
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addql #1,R(a0)
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cmpb MEM(a0),R(d0)
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beq L(L9)
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addql #1,R(a0)
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/* Otherwise the fourth byte must equal C. */
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L(L9:)
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movel R(a0),R(d0)
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movel MEM_POSTINC(sp),R(d3)
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movel MEM_POSTINC(sp),R(d2)
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rts
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END(__rawmemchr)
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libc_hidden_def (__rawmemchr)
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weak_alias (__rawmemchr, rawmemchr)
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