我正在尝试实现规范模式。在那,我有一个规范抽象类和一个重载的 && 运算符,它返回 AndSpecification,它实际上是从规范类派生的。现在我有循环依赖 b/w 的问题AndSpecification 和 Specification.Specification 是一个类模板,AndSpecification 也是。
规范.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;
};
将不胜感激。 完整代码:https://github.com/princekm/specification.git
答案 0 :(得分:1)
您需要前向声明,并在这些类之后移动需要两个类定义的方法定义:
// 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);
}
答案 1 :(得分:0)
转发声明:
template <class T> AndSpecification;
以下是正确的(通过删除示例中的纯虚方法):
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;
}
答案 2 :(得分:0)
您可以考虑 Specification
不一定需要了解 AndSpecification
,而不是前向声明(请参阅其他答案)。我的意思是,无论如何它都是一个模板。如果您像这样向 Specifiacation
添加模板模板参数:
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);
}
};
那么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>;