Question

假设我有两组关联类型，例如Animal s及其Offspring：

/* Animal types */
struct Animal
{
  virtual string getType() const = 0;
};

struct Cat : public Animal
{
  virtual string getType() const { return "Cat"; }
};

struct Dog : public Animal
{
  virtual string getType() const { return "Dog"; }
};


/* Offspring types */
struct Offspring
{
  virtual string getType() const = 0;
};

struct Kitten : public Offspring
{
  virtual string getType() const { return "Kitten"; }
};

struct Puppy : public Offspring
{
  virtual string getType() const { return "Puppy"; }
};

我正在尝试实施一个工厂，给定Animal将返回关联的Offspring类型的对象（例如，如果Animal实际上是Dog ，工厂将返回Puppy）。

我第一次尝试实施这样的工厂看起来像这样：

// First attempt at OffspringFactory
class OffspringFactory1
{
  static Offspring* createKitten() { return new Kitten(); }
  static Offspring* createPuppy()  { return new Puppy();  }

public:
  // Create an Offspring according to the Animal type
  static Offspring* getOffspring(const Animal& a)
  {
    // Static mapping of Animal types to Offspring factory functions
    static map<string, Offspring* (*)()> factoryMap;
    if (factoryMap.empty())
    {
      factoryMap["Dog"] = &createPuppy;
      factoryMap["Cat"] = &createKitten;
    }

    // Lookup our Offspring factory function
    map<string, Offspring* (*)()>::const_iterator fnIt = factoryMap.find(a.getType());
    if (fnIt != factoryMap.end())
      return fnIt->second();
    else
      throw "Bad animal type";
  }
};

它工作正常，但我使用的是基于字符串的映射，而不是纯粹基于类型的映射。在尝试转向更基于类型的实现时，我到达了这个：

// Second attempt at OffspringFactory
class OffspringFactory2
{
  // Mapping Animal types to Offspring types
  template<typename TAnimal> struct OffspringMapper;

  template<>
  struct OffspringMapper<Cat> {
    typedef Kitten offspring_type;
  };

  template<>
  struct OffspringMapper<Dog> {
    typedef Puppy offspring_type;
  };

  // Factory method
  template<typename TAnimal>
    static Offspring* create() { return new OffspringMapper<TAnimal>::offspring_type(); }

public:
  // Create an Offspring according to the Animal type
  static Offspring* getOffspring(const Animal& a)
  {
    // Static mapping of Animal type strings to Offspring factory functions
    static map<string, Offspring* (*)()> factoryMap;
    if (factoryMap.empty())
    {
      factoryMap["Dog"] = &create<Dog>;
      factoryMap["Cat"] = &create<Cat>;
    }

    // Lookup our Offspring factory function
    map<string, Offspring* (*)()>::const_iterator fnIt = factoryMap.find(a.getType());
    if (fnIt != factoryMap.end())
      return fnIt->second();
    else
      throw "Bad animal type";
  }
};

坦率地说，我不确定我在这里有什么改进：我仍然有我的字符串映射，以及更多不太可读的代码......

第二个实现是否有任何优点，并且有什么办法可以摆脱那个地图吗？

Answer 1

这看起来像是双重调度的经典案例。在C ++中解决此问题的一种模式是Visitor pattern。

class Offspring;
class OffspringFactory;

class Animal {
public:
    // ... rest of Animal class ...

    virtual Offspring* acceptOffspringFactory(OffspringFactory& factory)const = 0;
};

class OffspringFactory {
public:
    Offspring* createCatOffspring()
    {
        return new Kitten;
    }

    // ... one createXOffspring() for each type of Animal

    Offspring* getOffspring(const Animal& a)
    {
        return a.acceptOffspringFactory(*this);
    }
};

Offspring* Cat::acceptOffspringFactory(OffspringFactory& factory)const
{
    return factory.createCatOffspring();
}

// etc for rest of Animal classes

现在我再次查看你的问题，你没有表明工厂是抽象的，所以如果你能添加像@MooingDuck这样的方法，你真的可以完全取消工厂。

在对象工厂中键入映射

1 个答案: