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The following code fails to compile with type/value mismatch error, however I know a type is being supplied. What am I missing ?

template<typename A, typename B> struct base {}; template<typename B> struct derived : base< derived<B>::type, B > { using type = int; } int main() { derived<char> d; } error: type/value mismatch at argument 1 in template parameter list for 'template<class A, class B> struct base' struct derived : base< derived<B>::type, B > note: expected a type, got 'derived<B>::type'

Why is derived<B>::type not a valid type ?

Moreover tried the following :

template<typename B> struct derived : base< typename derived<B>::type, B > { using type = int; }

And got the following error :

no type name 'type' in 'struct derived<char>'

Why does compiler fail to detect type ?

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2 Answers 2

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You're seeing two different issues at play here. The first is that derived<B>::type is a dependent type name, so you have to use the typename keyword to inform the compiler that it is a type and not an object:

template<typename B> struct derived : base< typename derived<B>::type, B > { using type = int; }

That's not enough in this case though, since derived<B> isn't defined yet, so derived<B>::type doesn't exist yet.

For simple cases, you could just hard-code the type both places:

template<typename B> struct derived : base< int, B > { using type = int; }

For more complex cases, or if you want to avoid accidentally changing the template parameter and the using directive to be different in the future, you could pull type out into some other trait so that it's accessible before derived is defined:

template <typename B> struct my_trait { using type = int; }; template <typename B> struct derived : base < typename my_trait<B>::type, B > { using type = typename my_trait<B>::type; };

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Why does compiler fail to detect type ?

From class#mem-6

A class is considered a completely-defined object type ([basic.types]) (or complete type) at the closing } of the class-specifier.

Thus, struct derived is of incomplete type and declaring derived<B>::type as a template parameter would be ill-formed.

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