如何转发声明类模板？

Question

我试图实现规范pattern.In --我有一个规范抽象类和一个重载&操作符，它返回AndSpecification，它实际上是从class.Now派生的--存在循环依赖关系b/w AndSpecification的问题，Specification.Specification是类模板，AndSpecification也是。

Specification.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;
};

如能提供帮助，将不胜感激。完整代码：https://github.com/princekm/specification.git

Answer 1

您需要前向声明，并移动方法的定义，这两个类定义都需要在这些类之后：

// 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

向前声明：

template <class T> AndSpecification;

以下内容是正确的(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

与前向声明(请参阅其他答案)不同，您可以考虑Specification不一定需要了解AndSpecification。我是说，反正这是个模板。如果向Specifiacation添加模板模板参数，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，只有当您实例化它时，您才需要知道这两者：

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