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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 | /* $Id: umul.S,v 1.4 1996/09/30 02:22:39 davem Exp $ * umul.S: This routine was taken from glibc-1.09 and is covered * by the GNU Library General Public License Version 2. */ /* * Unsigned multiply. Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the * upper 32 bits of the 64-bit product). * * This code optimizes short (less than 13-bit) multiplies. Short * multiplies require 25 instruction cycles, and long ones require * 45 instruction cycles. * * On return, overflow has occurred (%o1 is not zero) if and only if * the Z condition code is clear, allowing, e.g., the following: * * call .umul * nop * bnz overflow (or tnz) */ .globl .umul .umul: or %o0, %o1, %o4 mov %o0, %y ! multiplier -> Y andncc %o4, 0xfff, %g0 ! test bits 12..31 of *both* args be Lmul_shortway ! if zero, can do it the short way andcc %g0, %g0, %o4 ! zero the partial product and clear N and V /* * Long multiply. 32 steps, followed by a final shift step. */ mulscc %o4, %o1, %o4 ! 1 mulscc %o4, %o1, %o4 ! 2 mulscc %o4, %o1, %o4 ! 3 mulscc %o4, %o1, %o4 ! 4 mulscc %o4, %o1, %o4 ! 5 mulscc %o4, %o1, %o4 ! 6 mulscc %o4, %o1, %o4 ! 7 mulscc %o4, %o1, %o4 ! 8 mulscc %o4, %o1, %o4 ! 9 mulscc %o4, %o1, %o4 ! 10 mulscc %o4, %o1, %o4 ! 11 mulscc %o4, %o1, %o4 ! 12 mulscc %o4, %o1, %o4 ! 13 mulscc %o4, %o1, %o4 ! 14 mulscc %o4, %o1, %o4 ! 15 mulscc %o4, %o1, %o4 ! 16 mulscc %o4, %o1, %o4 ! 17 mulscc %o4, %o1, %o4 ! 18 mulscc %o4, %o1, %o4 ! 19 mulscc %o4, %o1, %o4 ! 20 mulscc %o4, %o1, %o4 ! 21 mulscc %o4, %o1, %o4 ! 22 mulscc %o4, %o1, %o4 ! 23 mulscc %o4, %o1, %o4 ! 24 mulscc %o4, %o1, %o4 ! 25 mulscc %o4, %o1, %o4 ! 26 mulscc %o4, %o1, %o4 ! 27 mulscc %o4, %o1, %o4 ! 28 mulscc %o4, %o1, %o4 ! 29 mulscc %o4, %o1, %o4 ! 30 mulscc %o4, %o1, %o4 ! 31 mulscc %o4, %o1, %o4 ! 32 mulscc %o4, %g0, %o4 ! final shift /* * Normally, with the shift-and-add approach, if both numbers are * positive you get the correct result. With 32-bit two's-complement * numbers, -x is represented as * * x 32 * ( 2 - ------ ) mod 2 * 2 * 32 * 2 * * (the `mod 2' subtracts 1 from 1.bbbb). To avoid lots of 2^32s, * we can treat this as if the radix point were just to the left * of the sign bit (multiply by 2^32), and get * * -x = (2 - x) mod 2 * * Then, ignoring the `mod 2's for convenience: * * x * y = xy * -x * y = 2y - xy * x * -y = 2x - xy * -x * -y = 4 - 2x - 2y + xy * * For signed multiplies, we subtract (x << 32) from the partial * product to fix this problem for negative multipliers (see mul.s). * Because of the way the shift into the partial product is calculated * (N xor V), this term is automatically removed for the multiplicand, * so we don't have to adjust. * * But for unsigned multiplies, the high order bit wasn't a sign bit, * and the correction is wrong. So for unsigned multiplies where the * high order bit is one, we end up with xy - (y << 32). To fix it * we add y << 32. */ #if 0 tst %o1 bl,a 1f ! if %o1 < 0 (high order bit = 1), add %o4, %o0, %o4 ! %o4 += %o0 (add y to upper half) 1: rd %y, %o0 ! get lower half of product retl addcc %o4, %g0, %o1 ! put upper half in place and set Z for %o1==0 #else /* Faster code from tege@sics.se. */ sra %o1, 31, %o2 ! make mask from sign bit and %o0, %o2, %o2 ! %o2 = 0 or %o0, depending on sign of %o1 rd %y, %o0 ! get lower half of product retl addcc %o4, %o2, %o1 ! add compensation and put upper half in place #endif Lmul_shortway: /* * Short multiply. 12 steps, followed by a final shift step. * The resulting bits are off by 12 and (32-12) = 20 bit positions, * but there is no problem with %o0 being negative (unlike above), * and overflow is impossible (the answer is at most 24 bits long). */ mulscc %o4, %o1, %o4 ! 1 mulscc %o4, %o1, %o4 ! 2 mulscc %o4, %o1, %o4 ! 3 mulscc %o4, %o1, %o4 ! 4 mulscc %o4, %o1, %o4 ! 5 mulscc %o4, %o1, %o4 ! 6 mulscc %o4, %o1, %o4 ! 7 mulscc %o4, %o1, %o4 ! 8 mulscc %o4, %o1, %o4 ! 9 mulscc %o4, %o1, %o4 ! 10 mulscc %o4, %o1, %o4 ! 11 mulscc %o4, %o1, %o4 ! 12 mulscc %o4, %g0, %o4 ! final shift /* * %o4 has 20 of the bits that should be in the result; %y has * the bottom 12 (as %y's top 12). That is: * * %o4 %y * +----------------+----------------+ * | -12- | -20- | -12- | -20- | * +------(---------+------)---------+ * -----result----- * * The 12 bits of %o4 left of the `result' area are all zero; * in fact, all top 20 bits of %o4 are zero. */ rd %y, %o5 sll %o4, 12, %o0 ! shift middle bits left 12 srl %o5, 20, %o5 ! shift low bits right 20 or %o5, %o0, %o0 retl addcc %g0, %g0, %o1 ! %o1 = zero, and set Z .globl .umul_patch .umul_patch: umul %o0, %o1, %o0 retl rd %y, %o1 nop |