Bruno Haible <[EMAIL PROTECTED]> writes:
> Hi Ben,
>
> Paul Eggert wrote:
>> > * Is it acceptable for isfinite to raise an exception on
>> > a signaling NaN?
>>
>> Yes.
>
> I disagree. We know what the relevant standards say: [1]
> [1] http://lists.gnu.org/archive/html/bug-gnulib/2007-03/msg00396.html
Hmm, well, I checked into the standards. The C standard explicitly
says it doesn't define the behavior on signaling NaNs. (May 6, 2005
committee draft, F.2.1.)
The latest POSIX draft says this (section 4.20):
On implementations that support the IEC 60559: 1989 standard
floating point, functions with signaling NaN argument(s) shall be
treated as if the function were called with an argument that is a
required domain error and shall return a quiet NaN result, except
where stated otherwise.
Note: The function might never see the signaling NaN, since it
might trigger when the arguments are evaluated during the function
call.
The note suggests that, if sNaN is a signaling NaN, then
isfinite(sNaN) can trap even before isfinite starts executing, and
that the caller can't tell the difference between such a trap and
isfinite trapping. isfinite is a macro and not a function, but surely
the same idea applies to macros as well.
On implementations that support the IEC 60559: 1989 standard
floating point, for those functions that do not have a documented
domain error, the following shall apply:
These functions shall fail if:
Domain Error Any argument is a signaling NaN.
Either, the integer expression (math_errhandling &
MATH_ERRNO) is non-zero and errno shall be set to [EDOM], or
the integer expression (math_errhandling & MATH_ERREXCEPT) is
non-zero and the invalid floating-point exception shall be
raised.
Again, isfinite() is a macro and not a function; but isfinite() does
not have a documented domain error, so this spec suggests that
isfinite(sNaN) should either raise an exception, or should return 0
and set errno to EDOM.
Personally I agree with you that it's more convenient when isfinite
does not raise an exception. But the bottom line is that portable
code can't assume that isfinite(sNaN) will return 0 without raising an
exception.
