为模板参数类提供构造函数而无需继承

Question

I have a C++ class template. It takes a template parameter class T and stores an object of type T as a private member accessible through a method called data(). The code below demonstrates it. I'd like my class template to offer a convenient constructor which takes into account the type T. For example, a constructor which takes some of T's fields as parameters and initializes the encapsulated T object with their values.

One way to do that is having the user derive from the template instantiation and add their own constructors there, but I'd prefer, naturally, to have a way to supply ctors without making the user write a derived class.

template <class T>
class Templ
{
public:
     T& data();
     const T& data() const;
private:
     T obj;
};

Now if a user wants a convenienve constructor, they'd have to derive Templ:

class MyClass : public Templ<MyData>
{
public:
     MyClass (int size, MyClass* parent, float temperature, std::string name);
};

I read some C++11 stuff and had ideas about having a constructor template like the STL has std::list::emplace() set of methods, but I'm not sure what's the common best-practice solution.

Answer 1

Yeah, C++11 variadic perfect forwarding is what you want:

template <class T>
class Templ
{
public:
     template<typename... Args>
     explicit Templ(Args&&... args) : obj(std::forward<Args>(args)...) {}

     T& data();
     const T& data() const;
private:
     T obj;
};

This forwards any argument of the Templ constructor to the T constructor.

But I'm wondering, why even have this class at all? What are you trying to do?