Accessing base class template: is the following syntax legal?

[Vincent Reverdy]

Hello all,

Consider the following program, which compiles on both g++ and clang:

#include <iostream> 
template <class T> struct base {}; 
struct derived: base<int> {}; 
int main(int argc, char* argv[]) { 
    typename derived::base<double> x; // <- (1) Is this legal?
    typename derived::derived::base<double>::base<int>::base<void> y; // <- (2) Is this legal?
    return 0; 
}

Are both marked lines perfectly legal C++?

Thanks,
Vincent

[Brian Bi]

I think they are legal. See [temp.local]/1 (emphasis mine)

Like normal (non-template) classes, class templates have an injected-class-name (Clause 12). 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.

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Brian Bi

[Richard Smith]

The first one is legal. In the second one, I think derived::derived::base<double>::base<int>::base names a constructor, and so is ill-formed; see [class.qual]/2. Lookups of names followed by :: ignore function names, so "derived::derived::" does not name a constructor. But a constructor is a valid lookup result for (2), so that's ill-formed, I think.