Question

template <typename T>
class store  // Very basic class, capable of accepting any data-type and does nothing too much
{
  public:
    store(T value) : value(value) {}
  private:
    T value;
}

template <>
class store<int>  // Inherits all the basic functionality that the above class has and it also has additional methods
: public store<int> // PROBLEM OVER HERE. How do I refer to the above class?
{
  public:
    store(int value) : store<int>(value) /* PROBLEM OVER HERE. Should refer to the constructor of the above class */ {}
    void my_additional_int_method();
}

这里我有继承问题。我不想更改基类的名称，因为基类与所有派生类的用途相同（唯一的区别 - 派生类只有很少的额外方法）

Answer 1

你可以这样做：

template <typename T>
class store_impl
{
  public:
    store_impl(T value) : value(value) {}
  private:
    T value;
}

// default class accepting any type
// provides the default methods
template <typename T>
class store: public store_impl<T>
{
public:
    store(T value) : store_impl(value) {}
}

// specialization for int with extra methods
template <>
class store<int>: public store_impl<int>
{
  public:
    store(int value) : store_impl<int>(value)
    {}
    void my_additional_int_method();
}

Answer 2

您无法为类提供专用模板的名称：

template <>
class store<int>

你能做的就是给它一个具体的类型名称：

class store_int : public store<int>

或使用typedef或using声明

typdef store<int> store_int;

using store_int = store<int>;

模板类的继承

2 个答案: