Issue #14521: Make result of float('nan') and float('-nan') more consistent across platforms. Further, don't rely on Py_HUGE_VAL for float('inf').

This commit is contained in:
Mark Dickinson 2012-04-29 15:31:56 +01:00
parent d68ac85e9a
commit e383e82e04
5 changed files with 102 additions and 5 deletions

View file

@ -265,6 +265,16 @@ typedef union { double d; ULong L[2]; } U;
#define Big0 (Frac_mask1 | Exp_msk1*(DBL_MAX_EXP+Bias-1))
#define Big1 0xffffffff
/* Standard NaN used by _Py_dg_stdnan. */
#define NAN_WORD0 0x7ff80000
#define NAN_WORD1 0
/* Bits of the representation of positive infinity. */
#define POSINF_WORD0 0x7ff00000
#define POSINF_WORD1 0
/* struct BCinfo is used to pass information from _Py_dg_strtod to bigcomp */
typedef struct BCinfo BCinfo;
@ -1486,6 +1496,36 @@ bigcomp(U *rv, const char *s0, BCinfo *bc)
return 0;
}
/* Return a 'standard' NaN value.
There are exactly two quiet NaNs that don't arise by 'quieting' signaling
NaNs (see IEEE 754-2008, section 6.2.1). If sign == 0, return the one whose
sign bit is cleared. Otherwise, return the one whose sign bit is set.
*/
double
_Py_dg_stdnan(int sign)
{
U rv;
word0(&rv) = NAN_WORD0;
word1(&rv) = NAN_WORD1;
if (sign)
word0(&rv) |= Sign_bit;
return dval(&rv);
}
/* Return positive or negative infinity, according to the given sign (0 for
* positive infinity, 1 for negative infinity). */
double
_Py_dg_infinity(int sign)
{
U rv;
word0(&rv) = POSINF_WORD0;
word1(&rv) = POSINF_WORD1;
return sign ? -dval(&rv) : dval(&rv);
}
double
_Py_dg_strtod(const char *s00, char **se)
{
@ -1886,20 +1926,20 @@ _Py_dg_strtod(const char *s00, char **se)
bd2++;
/* At this stage bd5 - bb5 == e == bd2 - bb2 + bbe, bb2 - bs2 == 1,
and bs == 1, so:
and bs == 1, so:
tdv == bd * 10**e = bd * 2**(bbe - bb2 + bd2) * 5**(bd5 - bb5)
srv == bb * 2**bbe = bb * 2**(bbe - bb2 + bb2)
0.5 ulp(srv) == 2**(bbe-1) = bs * 2**(bbe - bb2 + bs2)
0.5 ulp(srv) == 2**(bbe-1) = bs * 2**(bbe - bb2 + bs2)
It follows that:
It follows that:
M * tdv = bd * 2**bd2 * 5**bd5
M * srv = bb * 2**bb2 * 5**bb5
M * 0.5 ulp(srv) = bs * 2**bs2 * 5**bb5
for some constant M. (Actually, M == 2**(bb2 - bbe) * 5**bb5, but
this fact is not needed below.)
for some constant M. (Actually, M == 2**(bb2 - bbe) * 5**bb5, but
this fact is not needed below.)
*/
/* Remove factor of 2**i, where i = min(bb2, bd2, bs2). */

View file

@ -22,6 +22,43 @@ case_insensitive_match(const char *s, const char *t)
the successfully parsed portion of the string. On failure, return -1.0 and
set *endptr to point to the start of the string. */
#ifndef PY_NO_SHORT_FLOAT_REPR
double
_Py_parse_inf_or_nan(const char *p, char **endptr)
{
double retval;
const char *s;
int negate = 0;
s = p;
if (*s == '-') {
negate = 1;
s++;
}
else if (*s == '+') {
s++;
}
if (case_insensitive_match(s, "inf")) {
s += 3;
if (case_insensitive_match(s, "inity"))
s += 5;
retval = _Py_dg_infinity(negate);
}
else if (case_insensitive_match(s, "nan")) {
s += 3;
retval = _Py_dg_stdnan(negate);
}
else {
s = p;
retval = -1.0;
}
*endptr = (char *)s;
return retval;
}
#else
double
_Py_parse_inf_or_nan(const char *p, char **endptr)
{
@ -57,6 +94,8 @@ _Py_parse_inf_or_nan(const char *p, char **endptr)
return retval;
}
#endif
/**
* _PyOS_ascii_strtod:
* @nptr: the string to convert to a numeric value.