Question

我一直在尝试通过使用C ++模板来实现访问者模式来减少代码中的样板量。到目前为止，我已经想出了这个：

class BaseVisitor {
public:
    virtual ~BaseVisitor() {}
};

template<typename T>
class Visitor : public BaseVisitor {
public:
    virtual void visit(T& /* visitable */) = 0;
};

template<typename Derived>
class Visitable {
public:
    void accept(Visitor<Derived>& visitor) {
        visitor.visit(static_cast<Derived&>(*this));
    }
};

Visitable的每个子类都是这样的：

class Mesh : public Object, public Visitable<Mesh> {};
class Text : public Object, public Visitable<Text> {};

最后，访客看起来像这样：

class Renderer : public Visitor<Mesh>, public Visitor<Text> {}

到目前为止一切都很好......现在问题在于：

for(Scene::iterator it = scene.begin(); it != scene.end(); ++it) {
    Object& object = static_cast<Object&>(*it);
    if(pre_visit(object)) {
        object.accept(this); ///Erm, what do I cast to??
        post_visit(object);
    }
}

我需要以某种方式转换为Visitable，以便我可以调用accept（），但显然我不知道T是什么。或者，我无法向Visitable模板添加虚拟accept（），因为我不知道应该采用什么参数。

任何C ++模板大师都知道如何使这项工作？

Answer 1

这可以使用可变参数模板在C ++ 11中完成。继续Pete的回答：

// Visitor template declaration
template<typename... Types>
class Visitor;

// specialization for single type    
template<typename T>
class Visitor<T> {
public:
    virtual void visit(T & visitable) = 0;
};

// specialization for multiple types
template<typename T, typename... Types>
class Visitor<T, Types...> : public Visitor<Types...> {
public:
    // promote the function(s) from the base class
    using Visitor<Types...>::visit;

    virtual void visit(T & visitable) = 0;
};

template<typename... Types>
class Visitable {
public:
    virtual void accept(Visitor<Types...>& visitor) = 0;
};

template<typename Derived, typename... Types>
class VisitableImpl : public Visitable<Types...> {
public:
    virtual void accept(Visitor<Types...>& visitor) {
        visitor.visit(static_cast<Derived&>(*this));
    }
};

Visitable的子类：

class Mesh : public Object, public VisitableImpl<Mesh, Mesh, Text> {};
class Text : public Object, public VisitableImpl<Text, Mesh, Text> {};

Visitor子类：

class Renderer : public Visitor<Mesh, Text> {};

目前尚不清楚value_type容器的Scene是什么，但您需要获取Visitable<Mesh, Text>的引用或指针accept：

for(Scene::iterator it = scene.begin(); it != scene.end(); ++it) {
    Visitable<Mesh, Text>& object = static_cast<Visitable<Mesh, Text>&>(*it);
    if(pre_visit(object)) {
        object.accept(*this);
        post_visit(object);
    }
}

Answer 2

除了允许任意访问者删除访问者之外，您的BaseVisitor不会为您做任何事情。相反，您希望为访问者提供一个基类，该基类提供可以在其上调用的不同accept函数的所有，并且Visitable接受此访问者

为此，您可以使用type list来定义访问者可以接受的类型，使用基类visitee类获取类型列表，并将类型列表作为参数添加到visitee实现中。< / p>

示例草图：

// assuming a typelist has typedefs first and second and a 
// type 'empty' representing end of type list

template<typename Types>
class Visitor : public Visitor<Types::second> {
public:
    // visitor has a visit function for each type in Types
    virtual void visit(typename Types::first& visitable) = 0;
};

template<> class Visitor<empty> { };

template<typename Types>
class Visitable{
    public:
    // base accepts a visitor which can visit any type in Types
    virtual void accept(Visitor<Types>& visitor) = 0;
};

template<typename Derived, typename Types>
class VisitableImpl : public Visitable<Types> {
public:
    // impl calls specific visit function 
    virtual void accept(Visitor<Types>& visitor) override {
        visitor.visit(static_cast<Derived&>(*this));
    }
};

Answer 3

我还需要模板化的Visitor模式，并且能够创建一个不涉及使用可变参数类型或类型列表的解决方案。

// forward declarations for our Visitable interface
class Object;
class Visitor;

// Visitable objects can accept a visitor.
class Visitable
{
public:
    virtual ~Visitable() { }
    virtual void accept_visitor(Visitor& visitor) = 0;
    virtual void accept(Object& obj);
};

// A base class, to allow downcasting
class Object
{
protected:
    virtual void _f() { }
};

// Our Visitor class, which will wrap our concrete visitor implementation
class Visitor
{
public:
    Visitor(Object* obj);

    // Base class for concrete visitors
    template<typename D, typename V>
    class OfType : public Object
    {
    public:
        void visit(V* visitable) {
            D* derived = static_cast<D*>(this);

            // "duck-typed" method; if our derived class does not have
            // this method, compilation will fail.
            derived->on_visit(visitable);
        }
    };

    template<typename D, typename V>
    void visit(V* visitable);

private:
    Object* m_obj;
};

Visitor::Visitor(Object* obj) : m_obj(obj) { }

template<typename D, typename V>
void Visitor::visit(V* visitable) {
    // check if our visitor is able to visit this instance
    OfType<D,V>* visitor = dynamic_cast<OfType<D,V>* >(m_obj);
    if (visitor) {
        visitor->visit(visitable);
    }
}

void Visitable::accept(Object& visitor) {
    Visitor wrapped(&visitor);
    accept_visitor(wrapped);
}

定义以上接口后，为可访问对象的访问者创建特定的接口，然后在您的具体类中实现它们：

class This;

class ThisVisitor : public Visitor::OfType<ThisVisitor, This>
{
public:
    virtual void on_visit(This* item) = 0;
};

class This : public Visitable
{
public:
    void accept_visitor(Visitor& visitor) {
        visitor.visit<ThisVisitor>(this);
    }
};

class That;

class ThatVisitor : public Visitor::OfType<ThatVisitor, That>
{
public:
    virtual void on_visit(That* item) = 0;
};

class That : public Visitable
{
public:
    void accept_visitor(Visitor& visitor) {
        visitor.visit<ThatVisitor>(this);
    }
};

class MyVisitor : public ThisVisitor, public ThatVisitor
{
public:
    void on_visit(This* item) { printf("This!"); }
    void on_visit(That* item) { printf("That!"); }
};

int main(int argc, const char* argv[] {
    This item1;
    That item2;
    MyVisitor visitor;
    item1.accept(visitor);   // "This!"
    item2.accept(visitor);   // "That!"
}

您还可以完全跳过访问者界面，直接让您的具体访问者从OfType<Derived, SomeClass>派生，但是我发现使用前者更好地扩展了您的访问者，因为定义了新类（That不应只要它是ThatVisitor类型，就可以关心谁访问它。

使用C ++模板实现访问者模式

3 个答案: