类型映射假设我有两套相关类型的,例如Animal
S和他们的Offspring
:在对象工厂
/* 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"; }
};
我想实现一个制造工厂,它给予Animal
将返回一个对象相关Offspring
类型(例如,如果Animal
事实上是Dog
,工厂将返回Puppy
)。
我在实现这样一个工厂第一次尝试是这样的:
// 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,但我已经使出了基于字符串的映射,而不是纯粹基于类型的事。在试图走向更基于类型的实现移动我来到这个:
// 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,我不知道我在这里的任何改进:更相当多的我还有我的字符串映射,沿可读性较差的代码行...
在第一次执行第二次执行时是否有任何优点,并且有什么方法可以摆脱该映射吗?
我不太确定你想达到什么目的?是否有某种你想要解决的现实生活问题? –
你可以添加'静态Offspring * Animal :: createOffspring()= 0;'?这将使这个_REALLY_变得容易。否则,你将不得不用一个枚举来替换你的字符串。 –
@MooingDuck我认为你的意思是'virtual'而不是static'在你的评论中。 –