I stand better informed than I really wanted to be, and only quacking quietly from getting my mind round the implications of these rather complex rules. (They remind me of PL/1 automatic conversions which were designed by IBM sales managers to sell computers but ended up being a maze of snares for the unwary.)
This scheme may offer more surprises to the many naive users (like me) than an explicit (and convenient 'global') choice, for example: using boost::math::double_constants; to ensure that the expected size is used. One can make the compiler warn you about size conversions, whereas I have the impression that these rules will mean that you won't get any warnings. Paul Paul A Bristow, Prizet Farmhouse, Kendal, Cumbria LA8 8AB UK +44 1539 561830 [EMAIL PROTECTED] | -----Original Message----- | From: [EMAIL PROTECTED] | [mailto:[EMAIL PROTECTED] On Behalf Of Gennaro Prota | Sent: 19 June 2003 11:10 | To: [EMAIL PROTECTED] | Subject: [boost] Re: Math Constants Formal Review - does | static_cast merely select? | | | On Wed, 18 Jun 2003 22:27:27 +0100, "Paul A Bristow" | <[EMAIL PROTECTED]> wrote: | | >| -----Original Message----- | >| From: [EMAIL PROTECTED] | >| [mailto:[EMAIL PROTECTED] On Behalf Of Gennaro Prota | >| [...] | >| | >| In Daniel's solution static_cast<type>(x) simply "selects" | >| the corresponding specialization of x_value. So | >| | >| static_cast<float>(pi) | >| | >| is fundamentally different from, say, | >| | >| double pi = ...; | >| static_cast<float>(pi) | > | >Perhaps naively, I find this surprising - but I am not a | language guru. | > | >Can you/anyone confirm that this is required by the Standard | > - and is in fact what happens when compilers process this? | | Yes :-) Note that all "conversions" from a constant<T, F> | pass through | | template< typename U > operator U() const { return F< U >()(); } | | Now, F<U>()() means (ignoring some technical differences irrelevant | here): | | F<U> object; // construct an x_value<U> object | return object(); // invoke operator() on it and return | | Now, if you use the binary operators, like in: | | pi * f; | | with f being a float then the selected specialization is | float; i.e. the constant "takes the type" of its adjacent | operand (left operand if it exists, right operand | otherwise). This is, of course, different from what normally | happens when you do | | double pi = ...; | pi * f; // floating point promotion on f | | and is IMHO a caveat. But, still, there are no truncations. | Only selection of the target specialization. If no target | specialization exists, e.g. | | static_cast<int>(pi) | | you have a compile-time error. | | > | >If it walks like a duck and quacks like a duck ... | | This walks quite differently. And doesn't quack :-) | | | Genny. | | _______________________________________________ | Unsubscribe & other changes: | http://lists.boost.org/mailman/listinfo.cgi/b| oost | _______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
