| /* $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 |