Question

Is it possible to use the ‘using’ declaration with template base classes? I have read it isn’t here but is that because of a technical reason or is it against the C++ standard, and does it apply to gcc or other compilers? If it is not possible, why not?

Example code (from the link above):

struct A {     template<class T> void f(T); };  struct B : A {     using A::f<int>; }; 

Answer 1

What you linked to is a using directive. A using declaration can be used fine with templated base classes (haven’t looked it up in the standard, but just tested it with a compiler):

template<typename T> struct c1 {      void foo() { std::cout << 'empty' << std::endl; }  };   template<typename T> struct c2 : c1<T> {      using c1<T>::foo;      void foo(int) { std::cout << 'int' << std::endl; }  };   int main() {      c2<void> c;     c.foo();     c.foo(10);  } 

The compiler correctly finds the parameter-less foo function because of our using-declaration re-declaring it into the scope of c2, and outputs the expected result.

Edit: updated the question. here is the updated answer:

The article is right about that you are not allowed to use a template-id (template name and arguments). But you can put a template name:

struct c1 {      template<int> void foo() { std::cout << 'empty' << std::endl; }  };   struct c2 : c1 {      using c1::foo; // using c1::foo<10> is not valid     void foo(int) { std::cout << 'int' << std::endl; }  };   int main() {      c2 c;     c.foo<10>();     c.foo(10);  }