4

我正在使用 CRTP 向继承的类添加克隆方法，例如：

class Base 
{
     virtual ~Base() {};
     virtual Base* clone() const = 0;
}; 

template<class Derived> class BaseCopyable : Base
{ 
public:
    virtual Base* clone() const
    {
        return new Derived(static_cast<Derived const&>(*this));
    }
};

class A : public BaseCopyable<A>;
class B : public BaseCopyable<B>;
etc...

但是如果我有一个继承自 B 的类，例如：

class differentB : public B;

那么clone()没有返回一个不同类型的对象，它返回一个B。除了在不同的B中写一个新的clone()方法，有没有办法解决这个问题？

谢谢阅读！

4

2 回答 2

4

这是我对这个问题的回答的返工

您的意图是让层次结构中的所有派生类从它们的基类继承可克隆性（多态副本），这样您就不需要为它们中的每一个提供覆盖clone()，但是您尝试使用类模板的 CRTP 解决方案 BaseCopyable只能以这种方式将可克隆性赋予直接从Base派生的类，而不是赋予从此类派生类派生的类。

我认为不可能通过在最顶层的具体类中“仅一次”授予可克隆性来将可克隆性传播到任意深度的层次结构。您必须将它明确地赋予每个具体类，但您可以通过它们的基类来做到这一点，而无需重复覆盖clone()，通过使用 CRTP 模板将可克隆性从父类传递到层次结构中的子类。

显然，符合此要求的 CRTP 模板与BaseCopyable 需要两个模板参数不同：父类型和子类型。

一个 C++03 解决方案如以下程序所示：

#include <iostream>

// As base of D, this makes D inherit B and makes D cloneable to
// a polymorphic pointer to B
template<class B, class D>
struct cloner : virtual B
{
    virtual B *clone() const {
        return new D(dynamic_cast<D const&>(*this));
    }
    virtual ~cloner() {}       
};

struct Base 
{
    virtual ~Base() {
         std::cout << "I was a Base" << std::endl;
    };
    virtual Base* clone() const = 0;
}; 

struct A : cloner<Base,A> // A inherits Base
{
    virtual ~A() {
         std::cout << "I was an A" << std::endl;
    };
};

struct B : cloner<Base,B> // B inherits Base
{
    virtual ~B() {
         std::cout << "I was a B" << std::endl;
    };
};

struct DB : cloner<B,DB> // DB inherits B, Base
{
    virtual ~DB() {
         std::cout << "I was a DB" << std::endl;
    };
};

int main()
{
    Base * pBaseA = new A;
    Base * pBaseB = new B;
    Base * pBaseDB = new DB;
    Base * pBaseCloneOfA = pBaseA->clone();
    Base * pBaseCloneOfB = pBaseB->clone();
    Base *pBaseCloneOfDB = pBaseDB->clone();
    B * pBCloneOfDB = dynamic_cast<B*>(pBaseDB->clone());
    std::cout << "deleting pBaseA" << std::endl; 
    delete pBaseA;
    std::cout << "deleting pBaseB" << std::endl;
    delete pBaseB;
    std::cout << "deleting pBaseDB" << std::endl;
    delete pBaseDB;
    std::cout << "deleting pBaseCloneOfA" << std::endl;
    delete pBaseCloneOfA;
    std::cout << "deleting pBaseCloneOfB" << std::endl; 
    delete pBaseCloneOfB;
    std::cout << "deleting pBaseCloneOfDB" << std::endl;    
    delete pBaseCloneOfDB;
    std::cout << "deleting pBCloneOfDB" << std::endl;   
    delete pBCloneOfDB;
    return 0;
}

输出是：

deleting pBaseA
I was an A
I was a Base
deleting pBaseB
I was a B
I was a Base
deleting pBaseDB
I was a DB
I was a B
I was a Base
deleting pBaseCloneOfA
I was an A
I was a Base
deleting pBaseCloneOfB
I was a B
I was a Base
deleting pBaseCloneOfDB
I was a DB
I was a B
I was a Base
deleting pBCloneOfDB
I was a DB
I was a B
I was a Base

假设所有涉及的类都是默认可构造的，B 不需要是的虚拟基础，cloner<B,D>并且您可以删除virtual 关键字 from struct cloner : virtual B。否则，B必须是虚拟基数，以便 的非默认构造函数B可以由 的构造函数调用D，尽管B不是 的直接基数D。

在 C++11 中，我们有可变参数模板，您可以通过提供“通用”模板构造函数来完全不使用虚拟继承，cloner<B,D>通过它可以将任意构造函数参数从 转发D到B. 这是一个例子：

#include <iostream>

template<class B, class D>
struct cloner : B
{
    B *clone() const override {
        return new D(dynamic_cast<D const&>(*this));
    }
    ~cloner() override {}
    // "All purpose constructor"
    template<typename... Args>
    explicit cloner(Args... args)
    : B(args...){}  
};

struct Base 
{
    explicit Base(int i)
    : _i(i){}   
    virtual ~Base() {
         std::cout << "I was a Base storing " << _i << std::endl;
    };
    virtual Base* clone() const = 0;
protected:
    int _i;
}; 

struct A : cloner<Base,A>
{
    explicit A(int i)
    : cloner<Base,A>(i){}
    ~A() override {
         std::cout << "I was an A storing " << _i << std::endl;
    };
};

struct B : cloner<Base,B>
{
    explicit B(int i)
    : cloner<Base,B>(i){}
    ~B() override {
         std::cout << "I was a B storing " << _i << std::endl;
    };
};

struct DB : cloner<B,DB>
{
    explicit DB(int i)
    : cloner<B,DB>(i){}
    ~DB() override {
         std::cout << "I was a DB storing " << _i << std::endl;
    };
};

int main()
{
    Base * pBaseA = new A(1);
    Base * pBaseB = new B(2);
    Base * pBaseDB = new DB(3);
    Base * pBaseCloneOfA = pBaseA->clone();
    Base * pBaseCloneOfB = pBaseB->clone();
    Base * pBaseCloneOfDB = pBaseDB->clone();
    B * pBCloneOfDB = dynamic_cast<B*>(pBaseDB->clone());
    std::cout << "deleting pA" << std::endl; 
    delete pBaseA;
    std::cout << "deleting pB" << std::endl;
    delete pBaseB;
    std::cout << "deleting pDB" << std::endl;
    delete pBaseDB;
    std::cout << "deleting pBaseCloneOfA" << std::endl;
    delete pBaseCloneOfA;
    std::cout << "deleting pBaseCloneOfB" << std::endl; 
    delete pBaseCloneOfB;
    std::cout << "deleting pBaseCloneOfDB" << std::endl;    
    delete pBaseCloneOfDB;
    std::cout << "deleting pBCloneOfDB" << std::endl;   
    delete pBCloneOfDB;
    return 0;
}

输出是：

deleting pA
I was an A storing 1
I was a Base storing 1
deleting pB
I was a B storing 2
I was a Base storing 2
deleting pDB
I was a DB storing 3
I was a B storing 3
I was a Base storing 3
deleting pBaseCloneOfA
I was an A storing 1
I was a Base storing 1
deleting pBaseCloneOfB
I was a B storing 2
I was a Base storing 2
deleting pBaseCloneOfDB
I was a DB storing 3
I was a B storing 3
I was a Base storing 3
deleting pBCloneOfDB
I was a DB storing 3
I was a B storing 3
I was a Base storing 3
于 2013-05-17T08:49:09.850 回答
0

您可以做的是通过整个继承层次结构传播基，但我认为这不会特别有用，因为对于每个进一步的派生类，您现在都会获得一个全新的层次结构，并且所有多态性都将化为乌有。

#include <iostream>

class Base 
{
public:
     virtual ~Base() {};
     virtual Base* clone() const = 0;
}; 

template<class Derived> class BaseCopyable : Base
{ 
public:
    virtual Base* clone() const
    {
        return new Derived(static_cast<Derived const&>(*this));
    }
};

struct Default;

template<typename Self, typename Arg>
struct SelfOrArg {
  typedef Arg type;
};

template<typename Self>
struct SelfOrArg<Self, Default> {
  typedef Self type;
};

template<typename Derived = Default>
class A : public BaseCopyable< typename SelfOrArg<A<Derived>, Derived>::type >
{

};

class derivedA : A<derivedA> {

};

虽然这仍然有坏的返回类型的缺点 BaseCopyable。使用经典virtual constructor成语，您可以说：

void func(Derived& d) {
  // thanks to covariant return types Derived::clone returns a Derived*
  Derived* d2 = d.clone();
  delete d2;
}

这在您的方案中是不可能的，尽管通过调整BaseCopyable.

只需编写一个宏来摆脱样板:)

于 2012-02-23T23:34:57.433 回答