Forwarding template parameters - c++

Question

I am trying to forward a template parameter to a parent class (from a subclass), however I am getting the following error when trying to do so;
test.cc:9:46: error: type/value mismatch at argument 1 in template parameter list for ‘template<class A, class B> class BaseClass’
class SubClass : public BaseClass<SubClass, B> {
^
test.cc:9:46: error: expected a type, got ‘SubClass’
Here is the code I am trying to run;
#include <cstdlib>
template<typename A, typename B>
class BaseClass {
};
template<typename B>
class SubClass : public BaseClass<SubClass, B> { // Trying to forward B to the second parameter
};
int main(void) {
return EXIT_SUCCESS;
}
I have tried prefixing B with typename, but I get the following error instead;
test.cc:9:55: error: type/value mismatch at argument 1 in template parameter list for ‘template<class A, class B> class BaseClass’
class SubClass : public BaseClass<SubClass, typename B> {
^
test.cc:9:55: error: expected a type, got ‘SubClass’
test.cc:9:55: error: template argument 2 is invalid

Answer 1

template<typename B>
class SubClass : public BaseClass<SubClass<B>, B>
// ^^^
{
};
SubClass is an injected name that can be used to refer to SubClass<B> only in the scope of SubClass. Outside, it is the name of a class template itself (that it, it requires a template parameter list <> to form a type).
§ 9 [class]/p2:
The class-name is also inserted into the scope of the class itself; this is known as the injected-class-name.
§ 3.3.2 [basic.scope.pdecl]/p8:
The point of declaration for an injected-class-name (Clause 9) is immediately following the opening brace of
the class definition.
§ 14.6.1 [temp.local]/p1:
Like normal (non-template) classes, class templates have an injected-class-name (Clause 9). The injected-class-name can be used as a template-name or a type-name. When it is used with a template-argument-list,
as a template-argument for a template template-parameter, or as the final identifier in the elaborated-type-specifier of a friend class template declaration, it refers to the class template itself. Otherwise, it is equivalent
to the template-name followed by the template-parameters of the class template enclosed in <>.

Answer 2

You're trying to use the injected-class-name. However,
The class-name is also inserted into the scope of the class itself;
this is known as the injected-class-name.
The base-clause is not looking up names that are within the scope of the class. Hence you have to supply the argument list yourself.
template<typename B>
class SubClass : public BaseClass<SubClass<B>, B> { /*..*/ };