将派生类成员函数指针存储在映射中

Question

我正在尝试为两个类Circle和Square实现一个工厂，它们都继承自Shape。

class Shape {
public: 
    virtual static
    Shape * getInstance() = 0;

};

class Circle : public Shape {        
public:
    static const std::string type;

    Shape * getInstance() {
        return new Circle;
    }
};
const std::string Circle::type = "Circle";

class Square : public Shape {        
public:
    static const std::string type;

    Shape * getInstance() {
        return new Square;
    }
};
const std::string Square::type = "Square"; 

我想创建一个以形状类型（字符串）为键，以相应派生类的getInstance()函数指针为值的映射表。这可行吗？

谢谢， Kiran

- Kiran Mohan

---

一个成员函数不能同时是static和virtual。你的类到底长什么样子？ - James McNellis

Answer 1

好的，我了解了错误。

1) 不应该在 Shape 类中声明 - virtual static Shape * getInstance() = 0;。

2) 在所有其他类中，getInstance() 应该是静态的。

这是完整的实现。

class Shape {
public:

    virtual
    std::string getType() = 0;

};

class Circle : public Shape {

public:
static const std::string type;
    Circle() {

    }

    std::string getType() {
        return type;
    }

    static
    Shape * getInstance() {
        return new Circle;
    }
};
const std::string Circle::type = "Circle";

class Square : public Shape {

public:
static const std::string type;
    Square() {
    }

    std::string getType() {
        return type;
    }

    static
    Shape * getInstance() {
        return new Square;
    }
};
const std::string Square::type = "Square";

class Triangle : public Shape {

public:
static const std::string type;
    Triangle() {
    }

    std::string getType() {
        return type;
    }

    static
    Shape * getInstance() {
        return new Triangle;
    }
};
const std::string Triangle::type = "Triangle";


typedef Shape * (*getShape)();
typedef std::map<std::string, getShape > factoryMap;

class ShapeFactory {
public:
    static factoryMap shapes;
    Shape * getInstance(const std::string & type){
        factoryMap::iterator itr = shapes.find(type);
        if (itr != shapes.end()){
            return (*itr->second)();
        }
        return NULL;
    }

};

factoryMap ShapeFactory::shapes;

class ShapeFactoryInitializer {
    static ShapeFactoryInitializer si;
public:

    ShapeFactoryInitializer() {
        ShapeFactory::shapes[Circle::type] = &Circle::getInstance;
        ShapeFactory::shapes[Square::type] = &Square::getInstance;
        ShapeFactory::shapes[Triangle::type] = &Triangle::getInstance;
    }

};

ShapeFactoryInitializer ShapeFactoryInitializer::si;

- Kiran Mohan

---

做了一些实验后，我意识到ShapeFactoryInitializer可能不太方便。比如你有一个新的形状叫做Polygon，你就必须找到ShapeFactoryIntializer并编辑它。相反，你可以这样做。 - Kiran Mohan

类 AddPentagonToShapeFactory{ private: static AddPentagonToShapeFactory addPolygonToShapeFactory; AddPentagonToShapeFactory(){ ShapeFactory::shapes[Pentagon::type] = &Pentagon::getInstance; } public: //无需公开任何内容 };AddPentagonToShapeFactory AddPentagonToShapeFactory::addPolygonToShapeFactory; - Kiran Mohan

当你自己解决了问题后，回答你的问题可以得到加1分。看到那些想要回馈社区的人真是太好了！ - A. Levy

Answer 2

虽然与您的问题关系不大，但如果您对现代C++设计（工厂、智能指针等）感兴趣，您可以查看这本书：

http://www.amazon.co.uk/Modern-Design-Applied-Generic-Patterns/dp/0201704315/ref=sr_1_20?s=books&ie=UTF8&qid=1293359949&sr=1-20

它谈论到工厂，如何设计它们等内容。

附注：我不是这本书的作者，也没有因发布这个答案而得到任何东西 :-)

- Rafid

Answer 3

将您的代码的最后一行更改为ShapeFactoryInitializer ShapeFactoryInitializer::si;，然后它就可以通过编译了。

- Answeror