diff options
| author | Jonathan Herman <hermanjl@cs.unc.edu> | 2013-01-22 10:38:37 -0500 |
|---|---|---|
| committer | Jonathan Herman <hermanjl@cs.unc.edu> | 2013-01-22 10:38:37 -0500 |
| commit | fcc9d2e5a6c89d22b8b773a64fb4ad21ac318446 (patch) | |
| tree | a57612d1888735a2ec7972891b68c1ac5ec8faea /arch/sparc/lib/umul.S | |
| parent | 8dea78da5cee153b8af9c07a2745f6c55057fe12 (diff) | |
Diffstat (limited to 'arch/sparc/lib/umul.S')
| -rw-r--r-- | arch/sparc/lib/umul.S | 171 |
1 files changed, 171 insertions, 0 deletions
diff --git a/arch/sparc/lib/umul.S b/arch/sparc/lib/umul.S new file mode 100644 index 00000000000..1f36ae68252 --- /dev/null +++ b/arch/sparc/lib/umul.S | |||
| @@ -0,0 +1,171 @@ | |||
| 1 | /* | ||
| 2 | * umul.S: This routine was taken from glibc-1.09 and is covered | ||
| 3 | * by the GNU Library General Public License Version 2. | ||
| 4 | */ | ||
| 5 | |||
| 6 | |||
| 7 | /* | ||
| 8 | * Unsigned multiply. Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the | ||
| 9 | * upper 32 bits of the 64-bit product). | ||
| 10 | * | ||
| 11 | * This code optimizes short (less than 13-bit) multiplies. Short | ||
| 12 | * multiplies require 25 instruction cycles, and long ones require | ||
| 13 | * 45 instruction cycles. | ||
| 14 | * | ||
| 15 | * On return, overflow has occurred (%o1 is not zero) if and only if | ||
| 16 | * the Z condition code is clear, allowing, e.g., the following: | ||
| 17 | * | ||
| 18 | * call .umul | ||
| 19 | * nop | ||
| 20 | * bnz overflow (or tnz) | ||
| 21 | */ | ||
| 22 | |||
| 23 | .globl .umul | ||
| 24 | .globl _Umul | ||
| 25 | .umul: | ||
| 26 | _Umul: /* needed for export */ | ||
| 27 | or %o0, %o1, %o4 | ||
| 28 | mov %o0, %y ! multiplier -> Y | ||
| 29 | |||
| 30 | andncc %o4, 0xfff, %g0 ! test bits 12..31 of *both* args | ||
| 31 | be Lmul_shortway ! if zero, can do it the short way | ||
| 32 | andcc %g0, %g0, %o4 ! zero the partial product and clear N and V | ||
| 33 | |||
| 34 | /* | ||
| 35 | * Long multiply. 32 steps, followed by a final shift step. | ||
| 36 | */ | ||
| 37 | mulscc %o4, %o1, %o4 ! 1 | ||
| 38 | mulscc %o4, %o1, %o4 ! 2 | ||
| 39 | mulscc %o4, %o1, %o4 ! 3 | ||
| 40 | mulscc %o4, %o1, %o4 ! 4 | ||
| 41 | mulscc %o4, %o1, %o4 ! 5 | ||
| 42 | mulscc %o4, %o1, %o4 ! 6 | ||
| 43 | mulscc %o4, %o1, %o4 ! 7 | ||
| 44 | mulscc %o4, %o1, %o4 ! 8 | ||
| 45 | mulscc %o4, %o1, %o4 ! 9 | ||
| 46 | mulscc %o4, %o1, %o4 ! 10 | ||
| 47 | mulscc %o4, %o1, %o4 ! 11 | ||
| 48 | mulscc %o4, %o1, %o4 ! 12 | ||
| 49 | mulscc %o4, %o1, %o4 ! 13 | ||
| 50 | mulscc %o4, %o1, %o4 ! 14 | ||
| 51 | mulscc %o4, %o1, %o4 ! 15 | ||
| 52 | mulscc %o4, %o1, %o4 ! 16 | ||
| 53 | mulscc %o4, %o1, %o4 ! 17 | ||
| 54 | mulscc %o4, %o1, %o4 ! 18 | ||
| 55 | mulscc %o4, %o1, %o4 ! 19 | ||
| 56 | mulscc %o4, %o1, %o4 ! 20 | ||
| 57 | mulscc %o4, %o1, %o4 ! 21 | ||
| 58 | mulscc %o4, %o1, %o4 ! 22 | ||
| 59 | mulscc %o4, %o1, %o4 ! 23 | ||
| 60 | mulscc %o4, %o1, %o4 ! 24 | ||
| 61 | mulscc %o4, %o1, %o4 ! 25 | ||
| 62 | mulscc %o4, %o1, %o4 ! 26 | ||
| 63 | mulscc %o4, %o1, %o4 ! 27 | ||
| 64 | mulscc %o4, %o1, %o4 ! 28 | ||
| 65 | mulscc %o4, %o1, %o4 ! 29 | ||
| 66 | mulscc %o4, %o1, %o4 ! 30 | ||
| 67 | mulscc %o4, %o1, %o4 ! 31 | ||
| 68 | mulscc %o4, %o1, %o4 ! 32 | ||
| 69 | mulscc %o4, %g0, %o4 ! final shift | ||
| 70 | |||
| 71 | |||
| 72 | /* | ||
| 73 | * Normally, with the shift-and-add approach, if both numbers are | ||
| 74 | * positive you get the correct result. With 32-bit two's-complement | ||
| 75 | * numbers, -x is represented as | ||
| 76 | * | ||
| 77 | * x 32 | ||
| 78 | * ( 2 - ------ ) mod 2 * 2 | ||
| 79 | * 32 | ||
| 80 | * 2 | ||
| 81 | * | ||
| 82 | * (the `mod 2' subtracts 1 from 1.bbbb). To avoid lots of 2^32s, | ||
| 83 | * we can treat this as if the radix point were just to the left | ||
| 84 | * of the sign bit (multiply by 2^32), and get | ||
| 85 | * | ||
| 86 | * -x = (2 - x) mod 2 | ||
| 87 | * | ||
| 88 | * Then, ignoring the `mod 2's for convenience: | ||
| 89 | * | ||
| 90 | * x * y = xy | ||
| 91 | * -x * y = 2y - xy | ||
| 92 | * x * -y = 2x - xy | ||
| 93 | * -x * -y = 4 - 2x - 2y + xy | ||
| 94 | * | ||
| 95 | * For signed multiplies, we subtract (x << 32) from the partial | ||
| 96 | * product to fix this problem for negative multipliers (see mul.s). | ||
| 97 | * Because of the way the shift into the partial product is calculated | ||
| 98 | * (N xor V), this term is automatically removed for the multiplicand, | ||
| 99 | * so we don't have to adjust. | ||
| 100 | * | ||
| 101 | * But for unsigned multiplies, the high order bit wasn't a sign bit, | ||
| 102 | * and the correction is wrong. So for unsigned multiplies where the | ||
| 103 | * high order bit is one, we end up with xy - (y << 32). To fix it | ||
| 104 | * we add y << 32. | ||
| 105 | */ | ||
| 106 | #if 0 | ||
| 107 | tst %o1 | ||
| 108 | bl,a 1f ! if %o1 < 0 (high order bit = 1), | ||
| 109 | add %o4, %o0, %o4 ! %o4 += %o0 (add y to upper half) | ||
| 110 | |||
| 111 | 1: | ||
| 112 | rd %y, %o0 ! get lower half of product | ||
| 113 | retl | ||
| 114 | addcc %o4, %g0, %o1 ! put upper half in place and set Z for %o1==0 | ||
| 115 | #else | ||
| 116 | /* Faster code from tege@sics.se. */ | ||
| 117 | sra %o1, 31, %o2 ! make mask from sign bit | ||
| 118 | and %o0, %o2, %o2 ! %o2 = 0 or %o0, depending on sign of %o1 | ||
| 119 | rd %y, %o0 ! get lower half of product | ||
| 120 | retl | ||
| 121 | addcc %o4, %o2, %o1 ! add compensation and put upper half in place | ||
| 122 | #endif | ||
| 123 | |||
| 124 | Lmul_shortway: | ||
| 125 | /* | ||
| 126 | * Short multiply. 12 steps, followed by a final shift step. | ||
| 127 | * The resulting bits are off by 12 and (32-12) = 20 bit positions, | ||
| 128 | * but there is no problem with %o0 being negative (unlike above), | ||
| 129 | * and overflow is impossible (the answer is at most 24 bits long). | ||
| 130 | */ | ||
| 131 | mulscc %o4, %o1, %o4 ! 1 | ||
| 132 | mulscc %o4, %o1, %o4 ! 2 | ||
| 133 | mulscc %o4, %o1, %o4 ! 3 | ||
| 134 | mulscc %o4, %o1, %o4 ! 4 | ||
| 135 | mulscc %o4, %o1, %o4 ! 5 | ||
| 136 | mulscc %o4, %o1, %o4 ! 6 | ||
| 137 | mulscc %o4, %o1, %o4 ! 7 | ||
| 138 | mulscc %o4, %o1, %o4 ! 8 | ||
| 139 | mulscc %o4, %o1, %o4 ! 9 | ||
| 140 | mulscc %o4, %o1, %o4 ! 10 | ||
| 141 | mulscc %o4, %o1, %o4 ! 11 | ||
| 142 | mulscc %o4, %o1, %o4 ! 12 | ||
| 143 | mulscc %o4, %g0, %o4 ! final shift | ||
| 144 | |||
| 145 | /* | ||
| 146 | * %o4 has 20 of the bits that should be in the result; %y has | ||
| 147 | * the bottom 12 (as %y's top 12). That is: | ||
| 148 | * | ||
| 149 | * %o4 %y | ||
| 150 | * +----------------+----------------+ | ||
| 151 | * | -12- | -20- | -12- | -20- | | ||
| 152 | * +------(---------+------)---------+ | ||
| 153 | * -----result----- | ||
| 154 | * | ||
| 155 | * The 12 bits of %o4 left of the `result' area are all zero; | ||
| 156 | * in fact, all top 20 bits of %o4 are zero. | ||
| 157 | */ | ||
| 158 | |||
| 159 | rd %y, %o5 | ||
| 160 | sll %o4, 12, %o0 ! shift middle bits left 12 | ||
| 161 | srl %o5, 20, %o5 ! shift low bits right 20 | ||
| 162 | or %o5, %o0, %o0 | ||
| 163 | retl | ||
| 164 | addcc %g0, %g0, %o1 ! %o1 = zero, and set Z | ||
| 165 | |||
| 166 | .globl .umul_patch | ||
| 167 | .umul_patch: | ||
| 168 | umul %o0, %o1, %o0 | ||
| 169 | retl | ||
| 170 | rd %y, %o1 | ||
| 171 | nop | ||
