1/* $NetBSD: umul.S,v 1.4 2003/08/07 16:42:23 agc Exp $ */ 2 3/* 4 * Copyright (c) 1992, 1993 5 * The Regents of the University of California. All rights reserved. 6 * 7 * This software was developed by the Computer Systems Engineering group 8 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 9 * contributed to Berkeley. 10 * 11 * Redistribution and use in source and binary forms, with or without 12 * modification, are permitted provided that the following conditions 13 * are met: 14 * 1. Redistributions of source code must retain the above copyright 15 * notice, this list of conditions and the following disclaimer. 16 * 2. Redistributions in binary form must reproduce the above copyright 17 * notice, this list of conditions and the following disclaimer in the 18 * documentation and/or other materials provided with the distribution. 19 * 3. Neither the name of the University nor the names of its contributors 20 * may be used to endorse or promote products derived from this software 21 * without specific prior written permission. 22 * 23 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 24 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 25 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 26 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 27 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 28 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 29 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 30 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 31 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 32 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 33 * SUCH DAMAGE. 34 * 35 * from: Header: umul.s,v 1.4 92/06/25 13:24:05 torek Exp 36 */ 37 38#include <machine/asm.h> 39#if defined(LIBC_SCCS) && !defined(lint) 40#if 0 41 .asciz "@(#)umul.s 8.1 (Berkeley) 6/4/93" 42#else 43 RCSID("$NetBSD: umul.S,v 1.4 2003/08/07 16:42:23 agc Exp $") 44#endif 45#endif /* LIBC_SCCS and not lint */ 46 47/* 48 * Unsigned multiply. Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the 49 * upper 32 bits of the 64-bit product). 50 * 51 * This code optimizes short (less than 13-bit) multiplies. Short 52 * multiplies require 25 instruction cycles, and long ones require 53 * 45 instruction cycles. 54 * 55 * On return, overflow has occurred (%o1 is not zero) if and only if 56 * the Z condition code is clear, allowing, e.g., the following: 57 * 58 * call .umul 59 * nop 60 * bnz overflow (or tnz) 61 */ 62 63FUNC(.umul) 64 or %o0, %o1, %o4 65 mov %o0, %y ! multiplier -> Y 66 andncc %o4, 0xfff, %g0 ! test bits 12..31 of *both* args 67 be Lmul_shortway ! if zero, can do it the short way 68 andcc %g0, %g0, %o4 ! zero the partial product and clear N and V 69 70 /* 71 * Long multiply. 32 steps, followed by a final shift step. 72 */ 73 mulscc %o4, %o1, %o4 ! 1 74 mulscc %o4, %o1, %o4 ! 2 75 mulscc %o4, %o1, %o4 ! 3 76 mulscc %o4, %o1, %o4 ! 4 77 mulscc %o4, %o1, %o4 ! 5 78 mulscc %o4, %o1, %o4 ! 6 79 mulscc %o4, %o1, %o4 ! 7 80 mulscc %o4, %o1, %o4 ! 8 81 mulscc %o4, %o1, %o4 ! 9 82 mulscc %o4, %o1, %o4 ! 10 83 mulscc %o4, %o1, %o4 ! 11 84 mulscc %o4, %o1, %o4 ! 12 85 mulscc %o4, %o1, %o4 ! 13 86 mulscc %o4, %o1, %o4 ! 14 87 mulscc %o4, %o1, %o4 ! 15 88 mulscc %o4, %o1, %o4 ! 16 89 mulscc %o4, %o1, %o4 ! 17 90 mulscc %o4, %o1, %o4 ! 18 91 mulscc %o4, %o1, %o4 ! 19 92 mulscc %o4, %o1, %o4 ! 20 93 mulscc %o4, %o1, %o4 ! 21 94 mulscc %o4, %o1, %o4 ! 22 95 mulscc %o4, %o1, %o4 ! 23 96 mulscc %o4, %o1, %o4 ! 24 97 mulscc %o4, %o1, %o4 ! 25 98 mulscc %o4, %o1, %o4 ! 26 99 mulscc %o4, %o1, %o4 ! 27 100 mulscc %o4, %o1, %o4 ! 28 101 mulscc %o4, %o1, %o4 ! 29 102 mulscc %o4, %o1, %o4 ! 30 103 mulscc %o4, %o1, %o4 ! 31 104 mulscc %o4, %o1, %o4 ! 32 105 mulscc %o4, %g0, %o4 ! final shift 106 107 108 /* 109 * Normally, with the shift-and-add approach, if both numbers are 110 * positive you get the correct result. WIth 32-bit two's-complement 111 * numbers, -x is represented as 112 * 113 * x 32 114 * ( 2 - ------ ) mod 2 * 2 115 * 32 116 * 2 117 * 118 * (the `mod 2' subtracts 1 from 1.bbbb). To avoid lots of 2^32s, 119 * we can treat this as if the radix point were just to the left 120 * of the sign bit (multiply by 2^32), and get 121 * 122 * -x = (2 - x) mod 2 123 * 124 * Then, ignoring the `mod 2's for convenience: 125 * 126 * x * y = xy 127 * -x * y = 2y - xy 128 * x * -y = 2x - xy 129 * -x * -y = 4 - 2x - 2y + xy 130 * 131 * For signed multiplies, we subtract (x << 32) from the partial 132 * product to fix this problem for negative multipliers (see mul.s). 133 * Because of the way the shift into the partial product is calculated 134 * (N xor V), this term is automatically removed for the multiplicand, 135 * so we don't have to adjust. 136 * 137 * But for unsigned multiplies, the high order bit wasn't a sign bit, 138 * and the correction is wrong. So for unsigned multiplies where the 139 * high order bit is one, we end up with xy - (y << 32). To fix it 140 * we add y << 32. 141 */ 142 tst %o1 143 bl,a 1f ! if %o1 < 0 (high order bit = 1), 144 add %o4, %o0, %o4 ! %o4 += %o0 (add y to upper half) 1451: rd %y, %o0 ! get lower half of product 146 retl 147 addcc %o4, %g0, %o1 ! put upper half in place and set Z for %o1==0 148 149Lmul_shortway: 150 /* 151 * Short multiply. 12 steps, followed by a final shift step. 152 * The resulting bits are off by 12 and (32-12) = 20 bit positions, 153 * but there is no problem with %o0 being negative (unlike above), 154 * and overflow is impossible (the answer is at most 24 bits long). 155 */ 156 mulscc %o4, %o1, %o4 ! 1 157 mulscc %o4, %o1, %o4 ! 2 158 mulscc %o4, %o1, %o4 ! 3 159 mulscc %o4, %o1, %o4 ! 4 160 mulscc %o4, %o1, %o4 ! 5 161 mulscc %o4, %o1, %o4 ! 6 162 mulscc %o4, %o1, %o4 ! 7 163 mulscc %o4, %o1, %o4 ! 8 164 mulscc %o4, %o1, %o4 ! 9 165 mulscc %o4, %o1, %o4 ! 10 166 mulscc %o4, %o1, %o4 ! 11 167 mulscc %o4, %o1, %o4 ! 12 168 mulscc %o4, %g0, %o4 ! final shift 169 170 /* 171 * %o4 has 20 of the bits that should be in the result; %y has 172 * the bottom 12 (as %y's top 12). That is: 173 * 174 * %o4 %y 175 * +----------------+----------------+ 176 * | -12- | -20- | -12- | -20- | 177 * +------(---------+------)---------+ 178 * -----result----- 179 * 180 * The 12 bits of %o4 left of the `result' area are all zero; 181 * in fact, all top 20 bits of %o4 are zero. 182 */ 183 184 rd %y, %o5 185 sll %o4, 12, %o0 ! shift middle bits left 12 186 srl %o5, 20, %o5 ! shift low bits right 20 187 or %o5, %o0, %o0 188 retl 189 addcc %g0, %g0, %o1 ! %o1 = zero, and set Z 190