1 /* @(#)e_fmod.c 1.3 95/01/18 */
2 /*-
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
5 *
6 * Developed at SunSoft, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
8 * software is freely granted, provided that this notice
9 * is preserved.
10 * ====================================================
11 */
12
13 #include <sys/cdefs.h>
14
15 #include "namespace.h"
16
17 #include <float.h>
18
19 #include "math.h"
20 #include "math_private.h"
21
22 #ifdef __weak_alias
23 __weak_alias(remquo, _remquo)
24 #endif
25
26 static const double Zero[] = {0.0, -0.0,};
27
28 /*
29 * Return the IEEE remainder and set *quo to the last n bits of the
30 * quotient, rounded to the nearest integer. We choose n=31 because
31 * we wind up computing all the integer bits of the quotient anyway as
32 * a side-effect of computing the remainder by the shift and subtract
33 * method. In practice, this is far more bits than are needed to use
34 * remquo in reduction algorithms.
35 */
36 double
remquo(double x,double y,int * quo)37 remquo(double x, double y, int *quo)
38 {
39 int32_t n,hx,hy,hz,ix,iy,sx,i;
40 u_int32_t lx,ly,lz,q,sxy;
41
42 EXTRACT_WORDS(hx,lx,x);
43 EXTRACT_WORDS(hy,ly,y);
44 sxy = (hx ^ hy) & 0x80000000;
45 sx = hx&0x80000000; /* sign of x */
46 hx ^=sx; /* |x| */
47 hy &= 0x7fffffff; /* |y| */
48
49 /* purge off exception values */
50 if((hy|ly)==0||(hx>=0x7ff00000)|| /* y=0,or x not finite */
51 ((hy|((ly|-ly)>>31))>0x7ff00000)) /* or y is NaN */
52 return (x*y)/(x*y);
53 if(hx<=hy) {
54 if((hx<hy)||(lx<ly)) {
55 q = 0;
56 goto fixup; /* |x|<|y| return x or x-y */
57 }
58 if(lx==ly) {
59 *quo = (sxy ? -1 : 1);
60 return Zero[(u_int32_t)sx>>31]; /* |x|=|y| return x*0*/
61 }
62 }
63
64 /* determine ix = ilogb(x) */
65 if(hx<0x00100000) { /* subnormal x */
66 if(hx==0) {
67 for (ix = -1043, i=lx; i>0; i<<=1) ix -=1;
68 } else {
69 for (ix = -1022,i=(hx<<11); i>0; i<<=1) ix -=1;
70 }
71 } else ix = (hx>>20)-1023;
72
73 /* determine iy = ilogb(y) */
74 if(hy<0x00100000) { /* subnormal y */
75 if(hy==0) {
76 for (iy = -1043, i=ly; i>0; i<<=1) iy -=1;
77 } else {
78 for (iy = -1022,i=(hy<<11); i>0; i<<=1) iy -=1;
79 }
80 } else iy = (hy>>20)-1023;
81
82 /* set up {hx,lx}, {hy,ly} and align y to x */
83 if(ix >= -1022)
84 hx = 0x00100000|(0x000fffff&hx);
85 else { /* subnormal x, shift x to normal */
86 n = -1022-ix;
87 if(n<=31) {
88 hx = (hx<<n)|(lx>>(32-n));
89 lx <<= n;
90 } else {
91 hx = lx<<(n-32);
92 lx = 0;
93 }
94 }
95 if(iy >= -1022)
96 hy = 0x00100000|(0x000fffff&hy);
97 else { /* subnormal y, shift y to normal */
98 n = -1022-iy;
99 if(n<=31) {
100 hy = (hy<<n)|(ly>>(32-n));
101 ly <<= n;
102 } else {
103 hy = ly<<(n-32);
104 ly = 0;
105 }
106 }
107
108 /* fix point fmod */
109 n = ix - iy;
110 q = 0;
111 while(n--) {
112 hz=hx-hy;lz=lx-ly; if(lx<ly) hz -= 1;
113 if(hz<0){hx = hx+hx+(lx>>31); lx = lx+lx;}
114 else {hx = hz+hz+(lz>>31); lx = lz+lz; q++;}
115 q <<= 1;
116 }
117 hz=hx-hy;lz=lx-ly; if(lx<ly) hz -= 1;
118 if(hz>=0) {hx=hz;lx=lz;q++;}
119
120 /* convert back to floating value and restore the sign */
121 if((hx|lx)==0) { /* return sign(x)*0 */
122 q &= 0x7fffffff;
123 *quo = (sxy ? -q : q);
124 return Zero[(u_int32_t)sx>>31];
125 }
126 while(hx<0x00100000) { /* normalize x */
127 hx = hx+hx+(lx>>31); lx = lx+lx;
128 iy -= 1;
129 }
130 if(iy>= -1022) { /* normalize output */
131 hx = ((hx-0x00100000)|((iy+1023)<<20));
132 } else { /* subnormal output */
133 n = -1022 - iy;
134 if(n<=20) {
135 lx = (lx>>n)|((u_int32_t)hx<<(32-n));
136 hx >>= n;
137 } else if (n<=31) {
138 lx = (hx<<(32-n))|(lx>>n); hx = 0;
139 } else {
140 lx = hx>>(n-32); hx = 0;
141 }
142 }
143 fixup:
144 INSERT_WORDS(x,hx,lx);
145 y = fabs(y);
146 if (y < 0x1p-1021) {
147 if (x+x>y || (x+x==y && (q & 1))) {
148 q++;
149 x-=y;
150 }
151 } else if (x>0.5*y || (x==0.5*y && (q & 1))) {
152 q++;
153 x-=y;
154 }
155 GET_HIGH_WORD(hx,x);
156 SET_HIGH_WORD(x,hx^sx);
157 q &= 0x7fffffff;
158 *quo = (sxy ? -q : q);
159 /*
160 * If q is 0 and we need to return negative, we have to choose
161 * the largest negative number (in 32 bits) because it is the
162 * only value that is negative and congruent to 0 mod 2^31.
163 */
164 if (q == 0 && sxy)
165 *quo = 0x80000000;
166 return x;
167 }
168