Bo Persson ha scritto:
From the proposal for new time classes (N2615)
http://www.open-std.org/jtc1/sc22/WG21/docs/papers/2008/n2615.html#common_type
I get this common_type traits
template <class ...T> struct common_type;
template <class T>
struct common_type<T>
{
typedef T type;
};
template <class T, class U>
struct common_type<T, U>
{
private:
static T&& t();
static U&& u();
public:
typedef decltype(true ? t() : u()) type;
};
Using gcc 4.3.1, which is the only compiler I have that compiles
it, I tried this simple code:typedef common_type<int>::type
one;typedef common_type<int, int>::type two;The compiler seems
to agree with me that 'one' is a typedef for 'int' while 'two' is
a typedef for 'int&&'. Why is that? Is that useful?Bo Persson
I believe it's a bug in gcc. According to my interpretation of
clause 5, paragraph 6, the type of the expression "true ? t() :
u()" cannot be a reference. In fact, the types of t() and u(),
which are T&& and U&&, are "adjusted" to T and U "prior to any
further analysis". Because of this adjustment the argument of the
conditional operator are simple rvalues (not rvalue-references!)
and there is nothing in 5.16 that might re-introduce the && in the
final type. (Before objecting to this interpretation, please be
aware of the subtle but fundamental difference between rvalue and
rvalue-reference.)
If my interpretation is correct, common_type<int, int>::type is just
int, as expected.
// ...
Could confuse anyone, I guess. :-)
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