Re: type traits technique
Olivier schrieb:
typedef char yes_type;
struct no_type { yes_type[2]; };
template< typename type >
struct is_class
{
yes_type test_class( void (type::*)( void ) );
no_type test_class( ... );
enum { res = (sizeof(yes_type) == sizeof(test_class(0))) };
};
this code basically implies that if the null pointer can be implicitly
casted to a class member function pointer taking no arguments and
returning no value, then 'type' is a class.
That is correct.
But why is that ? How can you be sure that a class will automatically
have a member function that corresponds to this signature 'void
(type::*)( void )' ? Is this suppose to match the constructor's
signature (i don't think that's true but I can't find an other
explaination). I've also seen other implementations using 'int
(type::*)( void ) as the signature match. Why is that ?
There is no relation to any constructor or other necessarily existing
member function. C++ does not require the existence of any member
function
or data member for the declaration and definition of a pointer to
function member or data member type, resp. This makes sense, because
there exist valid implicit and explicit conversions between pointer to
members along class hierachies. Generally the compiler cannot decide at
the point of declaration of such a pointer type whether there exists
(or could exist) such a derived class of a given one. The example 8.3.3
p. 2 clarifies this:
"[..] The declaration of pmd is well-formed even though X has no
members of type
double. Similarly, the declaration of pmc is well-formed even though Y
is an incomplete type. [..]" As you see there does even not need to
exist a class *definition* to define a pointer to member type of such
purely declared class.
Btw.: It is not possible to get the address of a constructor, see
12.1,p. 12: "[..] The address of a constructor shall not be taken."
Greetings from Bremen,
Daniel Kr?gler
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