On Sep 27, 10:45 pm, SzymonWlodar...@gmail.com wrote:
On Sep 27, 3:49 pm, chgans <chg...@googlemail.com> wrote:
I'm having difficulties with some template static member,
especially when this member is a template instance, for
example:
----
template<typename T>
class BaseT
{
public:
static void UseMap (const std::string &key, int value)
{
std::cout << gName << std::endl;
gMap[key] = value;
}
private:
static const std::string gName;
static std::map<std::string, int> gMap;
};
class DerivedT : public BaseT<DerivedT>
{
public:
// Some code soon or late....
};
// Now the specialization for BaseT<DerivedT>
// This one work fine
template<>
const std::string BaseT<DerivedT>::gName("Derived");
// This one gives me a linkage error:
// In function BaseT<DerivedT>::UseMap(...):
// undefined reference to BaseT<DerivedT>::gMap
template<>
std::map<std::string, int> BaseT<DerivedT>::gMap;
int main (int argc, char** argv)
{
DerivedT a;
a.UseMap ("test", 4);
}
----
So, i was wandering, if there is a special way to declare a
static member (which use the std::map template) of a
template.
It seems that if you specialize a static member you can't do
it with a default constructor. You can either write:
template< class T >
std::map<std::string, int> BaseT< T >::gMap;
or:
template<>
std::map<std::string, int> BaseT< DerivedT >::gMap( anotherMap );
However, I tested it using only one compiler (g++ 4.1.2) and I
did not look into the Standard so I am not sure that it is
what it requires.
Note that a specialization is not a template, but rather a
declaration or definition of a non-template entity with a name
that looks like a template instantiation, to be used instead of
the instantiation.
Givan that, the basic problem in this is that without an
initializer, the compiler interprets the static member
specialization as a declaration, not a definition, and since
it's not a template, you need a definition (in one, and only
one, translation unit). See =A714.7.3/15:
An explicit specialization of a static data member of a
template is a definition if the declaration includes an
initializer; otherwise, it is a declaration. [Note:
there is no syntax for the definition of a static data
member of a template which requires default
initialization.
template<> X Q<int>::x ;
This is a declaration regardless of whether X can be
default initialized.]
Note the note!
Note too that formally, you can only provide a single
definition, which means that the definition should be in a
source file, and not in a header; i.e.:
In the header:
template< class T >
std::map< std::string, int > BaseT< DerivedT >::gMap ;
and then in one and only one source file (which includes the
header):
template< class T >
std::map< std::string, int > BaseT< DerivedT >::gMap(
std::map< std::string, int > BaseT< DerivedT >() =