| /* $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 | |
| .globl _Umul | |
| .umul: | |
| _Umul: /* needed for export */ | |
| 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 |