Le lundi 15 mai 2006 à 16:22 +0100, Paul A Bristow a écrit :
Does this mean that the
// C99 macros defined as C++ templates template<class T> bool signbit(T x);
is undefined for a NaN?
No, it is perfectly defined as the value of the highest bit of the binary representation. But whether or not it is actually carrying any useful information is a whole other matter. NaN have sign bits but no sign. :)
I note that
ISO/IEC 9899:1999 (E) 7.19.6.1p8:
"A double argument representing an infinity is converted in one of the styles [-]inf or [-]infinity - which style is implementation-defined. A double argument representing a NaN is converted in one of the styles [-]nan or [-]nan(n-char-sequence)
Explicitly allows a preceeding - sign.
Seems like it is hard to justify prohibiting the - sign - but clearly its meaning is 'implementation defined'?
Right. My point was mainly about your sentence: "This allows for positive and negative infinities, and for both positive and negative quiet_NaN." I would simply avoid speaking of the sign for a NaN.
Similarly interpretation of the suffix (n-char-sequence) is too - but how is the stream expected to know when it has ended?
I may be wrong, but it seems to me the parentheses are present in the output, so the sequence trivially ends when the right parenthesis is reached. Best regards, Guillaume