This question is not duplicate but follow up of Propagating 'typedef' from based to derived class for 'template'
As a solution to inheriting of the typedefs, it was suggested to use using to import them to derived class, whereas simple using typename Base::typedefed_typeis supposed to suffice.
The following code is mostly taken from Roman Kruglov's answer:
#include <vector>
template<typename T>
class A
{
public:
typedef std::vector<T> Vec_t;
};
template<typename T>
class B : public A<T>
{
public:
using typename A::Vec_t;
// .........
private:
Vec_t v;
};
int main()
{
B<int> bb;
}
However it fails to compile, because compilers badly want template arguments of A.
Intel compiler error message:
1>C:\Work\EDPS\test_eigen\test_eigen.cpp(27): error : argument list for class template "A" is missing
1> using typename A::Vec_t;
1> ^
1> detected during instantiation of class "B<T> [with T=int]" at line 34
1>
1>C:\Work\EDPS\test_eigen\test_eigen.cpp(31): error : identifier "Vec_t" is undefined
1> Vec_t v;
1> ^
1> detected during instantiation of class "B<T> [with T=int]" at line 34
1>
MVC Error message:
c:\work\edps\test_eigen\test_eigen.cpp(27): error C2955: 'A': use of class template requires template argument list
1>c:\work\edps\test_eigen\test_eigen.cpp(17): note: see declaration of 'A'
1>c:\work\edps\test_eigen\test_eigen.cpp(32): note: see reference to class template instantiation 'B<T>' being compiled
1>c:\work\edps\test_eigen\test_eigen.cpp(27): error C3210: 'A': a member using-declaration can only be applied to a base class member
1>c:\work\edps\test_eigen\test_eigen.cpp(32): warning C4624: 'B<int>': destructor was implicitly defined as deleted
So what's wrong? Am I missing something? Or are perhaps the comments and answers there wrong??
I think the confusion arises from the way A is used as a base class. If a class template derives from a class template with a template argument, you have to fully qualify the base class name. But if a class derives from a class template specialization, you can use the base class name without a template argument list.
template<typename T>
struct A {
using t = T;
};
template<typename T>
struct B : A<T> {
using typename A<T>::t; // Full qualification needed -> mandatory argument list
};
struct C : A<int> {
using typename A::t; // Injected class name -> optional argument list
};
Also, note that t is available in C directly, without any using declaration or typedef. I may still be useful, for example if C inherits from A<int> privately, but you want t to be available publicly in C (in the example, C inherits publicly by default because it is a struct).