Re: template static member

James Kanze <>
Sun, 28 Sep 2008 00:23:14 -0700 (PDT)
On Sep 27, 10:45 pm, wrote:

On Sep 27, 3:49 pm, chgans <> wrote:

I'm having difficulties with some template static member,
especially when this member is a template instance, for
template<typename T>
class BaseT
  static void UseMap (const std::string &key, int value)
    std::cout << gName << std::endl;
    gMap[key] = value;

  static const std::string gName;
  static std::map<std::string, int> gMap;

class DerivedT : public BaseT<DerivedT>
   // Some code soon or late....

// Now the specialization for BaseT<DerivedT>

// This one work fine
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
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

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;


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

        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
    template< class T >
    std::map< std::string, int > BaseT< DerivedT >::gMap(
            std::map< std::string, int > BaseT< DerivedT >() ) ;
(Luckily, we can use the copy constructor in this case.)

James Kanze (GABI Software)
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