C ++:对于CRTP,在派生类中定义的类在基类中不可访问 [英] C++: With CRTP, class defined in the derived class is not accessible in the base class
问题描述
这里是(简化的)基类:
Here is the (simplified) base class:
template <class T>
class SharedObject
{
protected:
QExplicitlySharedDataPointer <typename T::Data> d;
};
这里是派生:
class ThisWontCompile : public SharedObject <ThisWontCompile>
{
private:
friend class SharedObject;
struct Data : public QSharedData
{
int id;
};
};
是否有任何解决方法可从访问 ThisWontCompile :: Data SharedObject ?
Is there any workaround to access ThisWontCompile::Data from SharedObject? What exactly can and what exactly cannot be done with the derived object from the base object?
推荐答案
这实际上是不相关的无障碍和友谊,它与CRTP的使用相关。考虑下面的示例也会出现问题:
This actually isn't related to the accessibility and friendship, it's related to the use of CRTP. Consider the following example that also exhibits the problem:
template <class T>
struct Base
{
typedef typename T::Data Data;
};
struct ThisWontCompile : public Base<ThisWontCompile>
{
struct Data { };
};
问题是 ThisWontCompile
它用作 Base
的模板参数的时间,因此它只能用作 Base
中的不完整类型。
The problem is that ThisWontCompile
is incomplete at the time it is used as a template argument to Base
, so it can only be used as an incomplete type in Base
.
有关您特定问题的一些解决方案,请参阅这个问题,特别是Martin的建议使用traits类,它基本上看起来像这样:
For a handful of solutions to your specific problem, consult the answers to this other question, especially Martin's recommendation to use a traits class, which would basically look like this:
// Base
template <typename T>
struct BaseTraits;
template <typename T>
struct Base
{
typedef typename BaseTraits<T>::Data Data;
};
// Derived
struct Derived;
template <>
struct BaseTraits<Derived>
{
struct Data { };
};
struct Derived : public Base<Derived>
{
};
typename BaseTraits< Derived> :: Data
可以在 Derived
和 Base
中使用。如果 Derived
本身是一个模板,您可以对traits类使用部分特化。
typename BaseTraits<Derived>::Data
can be used in both Derived
and in Base
. If Derived
is itself a template, you can use a partial specialization for the traits class.
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