Fix the conversion to use nan_mix() in r336362. fmod*(x, y),

remainder*(x, y) and remquo*(x, y, quo) were broken for y = 0 by changing
multiplication by y to addition of y.  (When y is 0, the result should be
NaN but became 1 for finite x.)

Use a new macro nan_mix_op() to give more control over the mixing, and
expand comments.

Recent re-testing missed finding this bug since I only tested the macro
version on amd64 and i386 and these arches don't use the C versions (they
use either asm versions or builtins).

Reported by:	enh via freebsd-numerics
This commit is contained in:
Bruce Evans 2018-07-24 10:10:16 +00:00
parent f9027e3a3a
commit daa1e39110
9 changed files with 23 additions and 16 deletions

View File

@ -42,7 +42,7 @@ __ieee754_fmod(double x, double y)
/* purge off exception values */
if((hy|ly)==0||(hx>=0x7ff00000)|| /* y=0,or x not finite */
((hy|((ly|-ly)>>31))>0x7ff00000)) /* or y is NaN */
return nan_mix(x, y)/nan_mix(x, y);
return nan_mix_op(x, y, *)/nan_mix_op(x, y, *);
if(hx<=hy) {
if((hx<hy)||(lx<ly)) return x; /* |x|<|y| return x */
if(lx==ly)

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@ -41,7 +41,7 @@ __ieee754_fmodf(float x, float y)
/* purge off exception values */
if(hy==0||(hx>=0x7f800000)|| /* y=0,or x not finite */
(hy>0x7f800000)) /* or y is NaN */
return nan_mix(x, y)/nan_mix(x, y);
return nan_mix_op(x, y, *)/nan_mix_op(x, y, *);
if(hx<hy) return x; /* |x|<|y| return x */
if(hx==hy)
return Zero[(u_int32_t)sx>>31]; /* |x|=|y| return x*0*/

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@ -79,7 +79,7 @@ fmodl(long double x, long double y)
(ux.bits.exp == BIAS + LDBL_MAX_EXP) || /* or x not finite */
(uy.bits.exp == BIAS + LDBL_MAX_EXP &&
((uy.bits.manh&~LDBL_NBIT)|uy.bits.manl)!=0)) /* or y is NaN */
return nan_mix(x, y)/nan_mix(x, y);
return nan_mix_op(x, y, *)/nan_mix_op(x, y, *);
if(ux.bits.exp<=uy.bits.exp) {
if((ux.bits.exp<uy.bits.exp) ||
(ux.bits.manh<=uy.bits.manh &&

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@ -49,7 +49,7 @@ __ieee754_remainder(double x, double p)
(hx>=0x7ff00000)|| /* x not finite */
((hp>=0x7ff00000)&& /* p is NaN */
(((hp-0x7ff00000)|lp)!=0)))
return nan_mix(x, p)/nan_mix(x, p);
return nan_mix_op(x, p, *)/nan_mix_op(x, p, *);
if (hp<=0x7fdfffff) x = __ieee754_fmod(x,p+p); /* now x < 2p */

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@ -39,7 +39,7 @@ __ieee754_remainderf(float x, float p)
if((hp==0)|| /* p = 0 */
(hx>=0x7f800000)|| /* x not finite */
((hp>0x7f800000))) /* p is NaN */
return nan_mix(x, p)/nan_mix(x, p);
return nan_mix_op(x, p, *)/nan_mix_op(x, p, *);
if (hp<=0x7effffff) x = __ieee754_fmodf(x,p+p); /* now x < 2p */

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@ -479,22 +479,29 @@ do { \
void _scan_nan(uint32_t *__words, int __num_words, const char *__s);
/*
* Mix 1 or 2 NaNs. First add 0 to each arg. This normally just turns
* Mix 0, 1 or 2 NaNs. First add 0 to each arg. This normally just turns
* signaling NaNs into quiet NaNs by setting a quiet bit. We do this
* because we want to never return a signaling NaN, and also because we
* don't want the quiet bit to affect the result. Then mix the converted
* args using addition. The result is typically the arg whose mantissa
* bits (considered as in integer) are largest.
* args using the specified operation.
*
* Technical complications: the result in bits might depend on the precision
* and/or on compiler optimizations, especially when different register sets
* are used for different precisions. Try to make the result not depend on
* at least the precision by always doing the main mixing step in long double
* When one arg is NaN, the result is typically that arg quieted. When both
* args are NaNs, the result is typically the quietening of the arg whose
* mantissa is largest after quietening. When neither arg is NaN, the
* result may be NaN because it is indeterminate, or finite for subsequent
* construction of a NaN as the indeterminate 0.0L/0.0L.
*
* Technical complications: the result in bits after rounding to the final
* precision might depend on the runtime precision and/or on compiler
* optimizations, especially when different register sets are used for
* different precisions. Try to make the result not depend on at least the
* runtime precision by always doing the main mixing step in long double
* precision. Try to reduce dependencies on optimizations by adding the
* the 0's in different precisions (unless everything is in long double
* precision).
*/
#define nan_mix(x, y) (((x) + 0.0L) + ((y) + 0))
#define nan_mix(x, y) (nan_mix_op((x), (y), +))
#define nan_mix_op(x, y, op) (((x) + 0.0L) op ((y) + 0))
#ifdef _COMPLEX_H

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@ -44,7 +44,7 @@ remquo(double x, double y, int *quo)
/* purge off exception values */
if((hy|ly)==0||(hx>=0x7ff00000)|| /* y=0,or x not finite */
((hy|((ly|-ly)>>31))>0x7ff00000)) /* or y is NaN */
return nan_mix(x, y)/nan_mix(x, y);
return nan_mix_op(x, y, *)/nan_mix_op(x, y, *);
if(hx<=hy) {
if((hx<hy)||(lx<ly)) {
q = 0;

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@ -41,7 +41,7 @@ remquof(float x, float y, int *quo)
/* purge off exception values */
if(hy==0||hx>=0x7f800000||hy>0x7f800000) /* y=0,NaN;or x not finite */
return nan_mix(x, y)/nan_mix(x, y);
return nan_mix_op(x, y, *)/nan_mix_op(x, y, *);
if(hx<hy) {
q = 0;
goto fixup; /* |x|<|y| return x or x-y */

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@ -86,7 +86,7 @@ remquol(long double x, long double y, int *quo)
(ux.bits.exp == BIAS + LDBL_MAX_EXP) || /* or x not finite */
(uy.bits.exp == BIAS + LDBL_MAX_EXP &&
((uy.bits.manh&~LDBL_NBIT)|uy.bits.manl)!=0)) /* or y is NaN */
return nan_mix(x, y)/nan_mix(x, y);
return nan_mix_op(x, y, *)/nan_mix_op(x, y, *);
if(ux.bits.exp<=uy.bits.exp) {
if((ux.bits.exp<uy.bits.exp) ||
(ux.bits.manh<=uy.bits.manh &&