SAS/C C++ Development System User's Guide, Release 6.50

Template Specialization Declarations

An explicit specialization is a declaration of a specialization of a template or template member that overrides generation of the specialization from its template definition. Such a specialization is called explicitly specialized. Members of explicitly specialized classes are treated as if they too are explicitly specialized.

Explicitly specialized items must be declared before their first use in every translation unit in which they are used. If the specialized item is used, it must be defined in a translation unit within the program.

The ANSI/ISO C++ draft form for an explicit specialization is a global declaration of the form:

template < > DECLARATION
The declaration refers to a specialization of a previously declared template or template member. The declaration may be a definition. If the declarator names a function template, an explicit template argument list may optionally follow the template name in order to disambiguate the specialization being declared. For example:

template <class T> class TypeAttr { public: static const int isIntegral; typedef T* Pointer; }; template <class T> const int TypeAttr<T>::isIntegral = 0; // default to // non-integral template < > const int TypeAttr<int>::isIntegral = 1; template < > const int TypeAttr<long>::isIntegral = 1; template <class T> T& deref( typename TypeAttr<T>::Pointer ); // T not // deducible template < > int& deref<int>( int *p ) // have to specify <int> // to disambiguate { return *p }

For compatibility with older compilers, if the TMPLFUNC option is not specified, ordinary function declarations that match the types of template specializations are treated as explicit specializations. As a special case, for compatibilty with older compilers, friend declarations inside a template class that use template parameter names in the declaration type are not treated as specializations. An example follows:

#include <iostream.h> template <class T> class Vector { // the following uses template // parameter "T", not a specialization friend ostream& operator<< ( ostream&, Vector<T>& ); // the following does not use // template parameters // It is treated as an explicit specialization friend ostream& operator<< ( ostream&, Vector<int>& ); . . . }; // define the template template <class T> ostream& operator<<( ostream& o, Vector<T>& v ) { . . . } // the following definition is also // treated as an explicit specialization ostream& operator<<( ostream& o, Vector<double>& v ) { . . . }

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