通过定义基类来“多态”声明非专业模板类型的成员的唯一方法是吗？

Question

Suppose we have a templated class,

template<typename Type>
class Container {};

Of course, we can't do this:

struct Foo
{
    Container _container;
}

But what if we wanted to do something like it? Is the only way to do this, to define a base class,

class ContainerBase {};

template<typename Type>
class Container : public ContainerBase {};

and store a pointer, like below?

struct Foo
{
    ContainerBase* _container;
}

It's simple enough, but it feels weird to have to add a base class solely for that reason, when it seems the compiler should have enough information to imply a set of related specializations. Of course, regardless _container needs to be a pointer, else Foo couldn't resolve to a static size, but

struct Foo
{
    Container* _container;
}

doesn't work either.

Answer 1

it seems the compiler should have enough information to imply a set of related specializations.

Nope. Template specializations are totally unrelated except in name, and the name of a type has essentially no bearing on runtime operation. Specializations of a given template usually share a (mostly) common interface, but they could just as well be completely different.

Adding a base class is essential if you want to relate between the specializations. And if they share so much in common, factoring that functionality into the base is a pretty great idea.