如何转发声明类模板

Question

我正在尝试实现规范模式。其中，我有一个规范抽象类和一个重载的&amp；返回AndSpecification的运算符，AndSpecification实际上是从Specification类派生的。现在我有一个循环依赖的问题，b/w AndSpecification和Specification。Specification是一个类模板，AndSpecification也是。

规格.h

//#include "AndSpecification.h"- this creates a problem
template <class T>
class Specification
{
public:
Specification()
{
}
virtual ~Specification()
{
}
virtual bool isSatisfied(T *item)=0;
AndSpecification<T>  operator&&(Specification &other)
{
return AndSpecification<T>(*this,other);
}
};

AndSpecification.h

#include "Specification.h"
template <class T>
class AndSpecification : public Specification<T>
{
Specification <T> *first;
Specification <T> *second;
public:
AndSpecification(Specification <T> *first,Specification <T> * second);
// Specification interface
public:
bool isSatisfied(T *item) override;
};

如有帮助，不胜感激。完整代码：https://github.com/princekm/specification.git

Answer 1

您需要前向声明，并将方法的定义移动到需要两个类定义的类之后：

// Specification.h
#pragma once
template <class T> class AndSpecification;
template <class T>
class Specification
{
public:
Specification() {}
virtual ~Specification() {}
virtual bool isSatisfied(T *item)=0;
AndSpecification<T>  operator&&(Specification &other);
};
#include "Specification.inl"

// AndSpecification.h
#pragma once
#include "Specification.h"
template <class T>
class AndSpecification : public Specification<T>
{
Specification <T> *first;
Specification <T> *second;
public:
AndSpecification(Specification <T> *first,Specification <T> * second);
// Specification interface
public:
bool isSatisfied(T *item) override;
};

// Specification.inl
#pragma once
#include "Specification.h"
#include "AndSpecification.h"
template <class T>
AndSpecification<T>  Specification<T>::operator&&(Specification &other)
{
return AndSpecification<T>(this, &other);
}

演示

Answer 2

转发声明：

template <class T> AndSpecification;

以下是正确的(通过删除示例中的纯虚拟方法(：

template <class T> class AndSpecification;
template <class T>
class Specification
{
public:
Specification()
{
}
virtual ~Specification()
{
}
//     virtual bool isSatisfied(T *item)=0;
AndSpecification<T>  operator&&(Specification &other)
{
return AndSpecification<T>(*this,other);
}
};
template <class T>
class AndSpecification : public Specification<T>
{
Specification <T> *first;
Specification <T> *second;
public:
AndSpecification(Specification <T> *first,Specification <T> * second);
// Specification interface
public:
bool isSatisfied(T *item) override;
};
class A{};
int main() {
Specification<A> a;
}

Answer 3

您可以考虑Specification不一定需要知道AndSpecification，而不是转发声明(请参阅其他答案(。我的意思是，不管怎样，这是一个模板。如果您将模板模板参数添加到Specifiacation中，如下所示：

template <class T,template<typename> class X>
class Specification
{
public:
Specification()
{
}
virtual ~Specification()
{
}
virtual bool isSatisfied(T *item)=0;
X<T>  operator&&(Specification &other)
{
return X<T>(*this,other);
}
};

然后只有当你实例化它时，你才需要知道两者：

#include "Specification.h"
#incldue "AndSpecification.h"
template <typename T>
using Specification_with_And = Specification<T,AndSpecification>;