NetBSD/sys/arch/sparc/fpu/fpu_div.c

266 lines
7.8 KiB
C

/* $NetBSD: fpu_div.c,v 1.5 2005/11/16 23:24:44 uwe Exp $ */
/*
* Copyright (c) 1992, 1993
* The Regents of the University of California. All rights reserved.
*
* This software was developed by the Computer Systems Engineering group
* at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
* contributed to Berkeley.
*
* All advertising materials mentioning features or use of this software
* must display the following acknowledgement:
* This product includes software developed by the University of
* California, Lawrence Berkeley Laboratory.
*
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions
* are met:
* 1. Redistributions of source code must retain the above copyright
* notice, this list of conditions and the following disclaimer.
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution.
* 3. Neither the name of the University nor the names of its contributors
* may be used to endorse or promote products derived from this software
* without specific prior written permission.
*
* THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
* ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
* SUCH DAMAGE.
*
* @(#)fpu_div.c 8.1 (Berkeley) 6/11/93
*/
/*
* Perform an FPU divide (return x / y).
*/
#include <sys/cdefs.h>
__KERNEL_RCSID(0, "$NetBSD: fpu_div.c,v 1.5 2005/11/16 23:24:44 uwe Exp $");
#include <sys/types.h>
#include <machine/reg.h>
#include <sparc/fpu/fpu_arith.h>
#include <sparc/fpu/fpu_emu.h>
/*
* Division of normal numbers is done as follows:
*
* x and y are floating point numbers, i.e., in the form 1.bbbb * 2^e.
* If X and Y are the mantissas (1.bbbb's), the quotient is then:
*
* q = (X / Y) * 2^((x exponent) - (y exponent))
*
* Since X and Y are both in [1.0,2.0), the quotient's mantissa (X / Y)
* will be in [0.5,2.0). Moreover, it will be less than 1.0 if and only
* if X < Y. In that case, it will have to be shifted left one bit to
* become a normal number, and the exponent decremented. Thus, the
* desired exponent is:
*
* left_shift = x->fp_mant < y->fp_mant;
* result_exp = x->fp_exp - y->fp_exp - left_shift;
*
* The quotient mantissa X/Y can then be computed one bit at a time
* using the following algorithm:
*
* Q = 0; -- Initial quotient.
* R = X; -- Initial remainder,
* if (left_shift) -- but fixed up in advance.
* R *= 2;
* for (bit = FP_NMANT; --bit >= 0; R *= 2) {
* if (R >= Y) {
* Q |= 1 << bit;
* R -= Y;
* }
* }
*
* The subtraction R -= Y always removes the uppermost bit from R (and
* can sometimes remove additional lower-order 1 bits); this proof is
* left to the reader.
*
* This loop correctly calculates the guard and round bits since they are
* included in the expanded internal representation. The sticky bit
* is to be set if and only if any other bits beyond guard and round
* would be set. From the above it is obvious that this is true if and
* only if the remainder R is nonzero when the loop terminates.
*
* Examining the loop above, we can see that the quotient Q is built
* one bit at a time ``from the top down''. This means that we can
* dispense with the multi-word arithmetic and just build it one word
* at a time, writing each result word when it is done.
*
* Furthermore, since X and Y are both in [1.0,2.0), we know that,
* initially, R >= Y. (Recall that, if X < Y, R is set to X * 2 and
* is therefore at in [2.0,4.0).) Thus Q is sure to have bit FP_NMANT-1
* set, and R can be set initially to either X - Y (when X >= Y) or
* 2X - Y (when X < Y). In addition, comparing R and Y is difficult,
* so we will simply calculate R - Y and see if that underflows.
* This leads to the following revised version of the algorithm:
*
* R = X;
* bit = FP_1;
* D = R - Y;
* if (D >= 0) {
* result_exp = x->fp_exp - y->fp_exp;
* R = D;
* q = bit;
* bit >>= 1;
* } else {
* result_exp = x->fp_exp - y->fp_exp - 1;
* q = 0;
* }
* R <<= 1;
* do {
* D = R - Y;
* if (D >= 0) {
* q |= bit;
* R = D;
* }
* R <<= 1;
* } while ((bit >>= 1) != 0);
* Q[0] = q;
* for (i = 1; i < 4; i++) {
* q = 0, bit = 1 << 31;
* do {
* D = R - Y;
* if (D >= 0) {
* q |= bit;
* R = D;
* }
* R <<= 1;
* } while ((bit >>= 1) != 0);
* Q[i] = q;
* }
*
* This can be refined just a bit further by moving the `R <<= 1'
* calculations to the front of the do-loops and eliding the first one.
* The process can be terminated immediately whenever R becomes 0, but
* this is relatively rare, and we do not bother.
*/
struct fpn *
fpu_div(struct fpemu *fe)
{
register struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2;
register u_int q, bit;
register u_int r0, r1, r2, r3, d0, d1, d2, d3, y0, y1, y2, y3;
FPU_DECL_CARRY
/*
* Since divide is not commutative, we cannot just use ORDER.
* Check either operand for NaN first; if there is at least one,
* order the signalling one (if only one) onto the right, then
* return it. Otherwise we have the following cases:
*
* Inf / Inf = NaN, plus NV exception
* Inf / num = Inf [i.e., return x]
* Inf / 0 = Inf [i.e., return x]
* 0 / Inf = 0 [i.e., return x]
* 0 / num = 0 [i.e., return x]
* 0 / 0 = NaN, plus NV exception
* num / Inf = 0
* num / num = num (do the divide)
* num / 0 = Inf, plus DZ exception
*/
if (ISNAN(x) || ISNAN(y)) {
ORDER(x, y);
return (y);
}
if (ISINF(x) || ISZERO(x)) {
if (x->fp_class == y->fp_class)
return (fpu_newnan(fe));
return (x);
}
/* all results at this point use XOR of operand signs */
x->fp_sign ^= y->fp_sign;
if (ISINF(y)) {
x->fp_class = FPC_ZERO;
return (x);
}
if (ISZERO(y)) {
fe->fe_cx = FSR_DZ;
x->fp_class = FPC_INF;
return (x);
}
/*
* Macros for the divide. See comments at top for algorithm.
* Note that we expand R, D, and Y here.
*/
#define SUBTRACT /* D = R - Y */ \
FPU_SUBS(d3, r3, y3); FPU_SUBCS(d2, r2, y2); \
FPU_SUBCS(d1, r1, y1); FPU_SUBC(d0, r0, y0)
#define NONNEGATIVE /* D >= 0 */ \
((int)d0 >= 0)
#ifdef FPU_SHL1_BY_ADD
#define SHL1 /* R <<= 1 */ \
FPU_ADDS(r3, r3, r3); FPU_ADDCS(r2, r2, r2); \
FPU_ADDCS(r1, r1, r1); FPU_ADDC(r0, r0, r0)
#else
#define SHL1 \
r0 = (r0 << 1) | (r1 >> 31), r1 = (r1 << 1) | (r2 >> 31), \
r2 = (r2 << 1) | (r3 >> 31), r3 <<= 1
#endif
#define LOOP /* do ... while (bit >>= 1) */ \
do { \
SHL1; \
SUBTRACT; \
if (NONNEGATIVE) { \
q |= bit; \
r0 = d0, r1 = d1, r2 = d2, r3 = d3; \
} \
} while ((bit >>= 1) != 0)
#define WORD(r, i) /* calculate r->fp_mant[i] */ \
q = 0; \
bit = 1 << 31; \
LOOP; \
(x)->fp_mant[i] = q
/* Setup. Note that we put our result in x. */
r0 = x->fp_mant[0];
r1 = x->fp_mant[1];
r2 = x->fp_mant[2];
r3 = x->fp_mant[3];
y0 = y->fp_mant[0];
y1 = y->fp_mant[1];
y2 = y->fp_mant[2];
y3 = y->fp_mant[3];
bit = FP_1;
SUBTRACT;
if (NONNEGATIVE) {
x->fp_exp -= y->fp_exp;
r0 = d0, r1 = d1, r2 = d2, r3 = d3;
q = bit;
bit >>= 1;
} else {
x->fp_exp -= y->fp_exp + 1;
q = 0;
}
LOOP;
x->fp_mant[0] = q;
WORD(x, 1);
WORD(x, 2);
WORD(x, 3);
x->fp_sticky = r0 | r1 | r2 | r3;
return (x);
}