1/*- 2 * Copyright (c) 2013 The NetBSD Foundation, Inc. 3 * All rights reserved. 4 * 5 * This code is derived from software contributed to The NetBSD Foundation 6 * by Matt Thomas of 3am Software Foundry. 7 * 8 * Redistribution and use in source and binary forms, with or without 9 * modification, are permitted provided that the following conditions 10 * are met: 11 * 1. Redistributions of source code must retain the above copyright 12 * notice, this list of conditions and the following disclaimer. 13 * 2. Redistributions in binary form must reproduce the above copyright 14 * notice, this list of conditions and the following disclaimer in the 15 * documentation and/or other materials provided with the distribution. 16 * 17 * THIS SOFTWARE IS PROVIDED BY THE NETBSD FOUNDATION, INC. AND CONTRIBUTORS 18 * ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED 19 * TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR 20 * PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE FOUNDATION OR CONTRIBUTORS 21 * BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR 22 * CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF 23 * SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS 24 * INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN 25 * CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) 26 * ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE 27 * POSSIBILITY OF SUCH DAMAGE. 28 */ 29 30#include <machine/asm.h> 31 32RCSID("$NetBSD: memset_naive.S,v 1.1 2013/01/08 20:15:00 matt Exp $") 33 34/* 35 * This isn't quite as simple/short as it could be but the truly trivial 36 * memset was an order of magnitude slower than this. 37 */ 38 39ENTRY(memset) 40/* LINTSTUB: void *memset(void *, int, size_t) */ 41 mov ip, r0 /* need to preserve r0 */ 42 cmp r2, #10 /* 10 bytes or less? */ 43 ble .Lbyte_by_byte /* yes, bytewise is faster */ 44 ands r3, r1, #0xff /* we are dealing with bytes */ 45 orrne r3, r3, r3, lsl #8 /* move value into 2nd byte lane */ 46 orrne r3, r3, r3, lsl #16 /* move value into all byte lanes */ 47 mov r1, r2 /* move count */ 48 ands r2, ip, #7 /* are we dword aligned? */ 49 beq 1f /* yes we are */ 50 rsb r2, r2, #8 /* how many bytes until aligned? */ 51 sub r1, r1, r2 /* subtract from count */ 52 tst ip, #1 /* halfword aligned? */ 53 strneb r3, [ip], #1 /* nope, write a byte */ 54 tst ip, #2 /* word aligned? */ 55 strneh r3, [ip], #2 /* nope, write a halfword */ 56 tst ip, #4 /* dword aligned? */ 57 strne r3, [ip], #4 /* nope, write a word */ 58 /* 59 * At this point, we are dword aligned. 60 */ 611: mov r2, r3 /* duplicate fill value */ 622: subs r1, r1, #16 /* can we write 16 bytes? */ 63 stmgeia ip!, {r2,r3} /* yes, write the first 8 of them */ 64 stmgeia ip!, {r2,r3} /* yes, write the second 8 of them */ 65 bgt 2b /* more left to fill */ 66 RETc(eq) /* no, return */ 67 /* 68 * Our count went negative but the bits below 16 haven't changed. 69 * So we are effectively testing modulo 16. 70 */ 71 tst r1, #8 /* can we write at least 8 bytes? */ 72 stmneia ip!, {r2,r3} /* so do it */ 73 tst r1, #4 /* can we write at least 4 bytes? */ 74 strne r3, [ip], #4 /* so do it */ 75 tst r1, #2 /* can we write at least 2 bytes? */ 76 strneh r3, [ip], #2 /* so do it */ 77 tst r1, #1 /* can we write 1 bytes? */ 78 strneb r3, [ip], #1 /* so do it */ 79 RET /* return */ 80 81.Lbyte_by_byte: 82 subs r2, r2, #1 /* can we write a byte? */ 83 RETc(lt) /* no, return */ 84 strb r3, [ip], #1 /* write a byte */ 85 b .Lbyte_by_byte /* do next byte */ 86END(memset) 87