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Asked: 2026-06-12T18:33:44+00:00

Here’s the scenario: template <template <typename> class T, typename V> struct parent { void

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Here’s the scenario:

template <template <typename> class T, typename V>
struct parent {
    void do_something();
};

template <typename V>
struct child : public parent<child, V> {
    void do_something(V argument);
    using parent<child, V>::do_something; // C3200: invalid template argument for template parameter 'IMPL', expected a class template
};

The above code fails to compile on the given line with the given error (MSVC 9.0). However if I write this instead, outside of the class definition for child:

template <typename V>
struct parent_identity_meta {
    typedef typename parent<child, V> type; // no error!
};

I can now successfully do the following, within child:

using parent_identity_meta<V>::type::do_something;

I know there’s a limitation (alleviated in C++11) that you can’t typedef against a template, but I don’t think that’s what I’m running into here, otherwise the typedef in parent_identity_meta would fail. It seems like child refers to the template when not inside of its own class definition, and to the class being generated from within itself.

This is pretty understandable (having to write child<V> every single time would be painful); but is there any way to override this behaviour?

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1 Answer

  1. Editorial Team

    This is a place where C++03 and C++11 are different from each other. The relevant part of the standard is [temp.local]/1. In C++03, this states:

    Like normal (non-template) classes, class templates have an injected-class-name (clause 9). The injected-class-name can be used with or without a template-argument-list. When it is used without a template- argument-list, it is equivalent to the injected-class-name followed by the template-parameters of the class template enclosed in <>. When it is used with a template-argument-list, it refers to the specified class template specialization, which could be the current specialization or another specialization.

    This means that child (without any template arguments) refers to the specialization child<V>. In C++11, it was changed to:

    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 <>.

    Note in particular When it is used ... as a template-argument for a template template-parameter ... it refers to the class template itself.. This means that in C++11, your code would be correct.

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