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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').

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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
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50
Python/dtoa.c
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@ -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). */