Why must we specify template arguments for template base class when calling its constructor? [duplicate]

Question

I was porting some code from MSVC(without permissive-) to linux and I learned that if you call constructor of a template base class in initializer list of your class you must specify all the template parameters or you get an error. Seems kind of redundant, since if you make a mistake in retyping the template parameters it is a hard error:

error: type 'Base<int, true>' is not a direct or virtual base of 'Derived'

full code here:

template <typename T, bool has_x>
struct Base
{
    Base(T t): t_(t){
    }
    T t_=0;
};



template <typename T>
class Derived : public Base<T, false>
{
public:
    // : Base<T, true> is hard error
    Derived(const T& t) : Base<T, false>(t) {}
};

int main()
{
    Derived d(47);
}

Is there a strong reason for this, or just standardization process never took time to special case this use case?

Answer 1

You only need to do that when Derived is a template and the type of the base depends on its template parameters.

This compiles, for example:

template <typename T>
class Derived : public Base<int, false>
{
public:
    Derived(const T& t) : Base(t) {}
};

As far as I know, here (in member initializer list) Base is actually the injected-class-name of Base<...>, inherited from it like everything else.

And if the type of the base does depend on the template parameters, its inherited injected-class-name becomes inaccessible (at least directly), just like any other member inherited from it.

For a member variable/function, you'd add this-> to access it, but for a type member you need Derived:::

template <typename T>
class Derived : public Base<T, false>
{
public:
    Derived(const T& t) : Derived::Base(t) {}
};