我正在尝试创建一个管理器,它通过模板提供对存储不同类型数据的容器的访问。
首先,我制作了用于存储不同类型数据的基本管理器类
template <typename T>
class BaseMapManager{
public:
T add(T element, std::string name);
T remove(T element);
T remove(std::string name);
T remove(unsigned int id);
T get(std::string name);
T get(unsigned int id);
BaseMapManager() { id = 0; };
~BaseMapManager() { nameMap.clear(); idMap.clear(); };
protected:
std::map<std::string, T> nameMap;
std::map<unsigned int, T> idMap;
//each element of type T gets new unique id
unsigned int id;
//hide it
BaseMapManager(const BaseMapManager&) {};
BaseMapManager& operator=(const BaseMapManager&) {};
};
然后我创建了我的具体管理器,它必须存储 3 种类型的 baseMapManager:
class ResourceManager{
public:
//creates singleton
static ResourceManager* init(){
static ResourceManager singleton;
return &singleton;
}
template <typename T>
void load_resource(std::string path, std::string name){
get_map<std::shared_ptr<T>>().add(std::shared_ptr<T>(new T(path, name), name);
}
template <typename T>
std::shared_ptr<T> get_resource(std::string name){
get_map<T>().get(name);
}
template <typename T>
std::shared_ptr<T> get_resource(unsigned int id){
get_map<T>().get(id);
}
private:
BaseMapManager<std::shared_ptr<AnimationResource> > animationMap;
BaseMapManager<std::shared_ptr<ImageResource> > imageMap;
BaseMapManager<std::shared_ptr<FontResource> > fontMap;
template <typename T>
BaseMapManager<T>& get_map(){
if (std::is_same<T, std::shared_ptr<AnimationResource> >() == true) return animationMap;
if (std::is_same<T, std::shared_ptr<ImageResource> >() == true) return imageMap;
if (std::is_same<T, std::shared_ptr<FontResource> >() == true) return fontMap;
};
};
现在我得到了这个:
1>------ Build started: Project: BOSS, Configuration: Debug Win32 ------
1> main.cpp
1>d:\programming\github projects\boss\boss\new\resourcemanager\resourcemanager.h(43):
error C2440: 'return' : cannot convert from 'BaseMapManager<T>' to 'BaseMapManager<T> &'
1> with
1> [
1> T=std::shared_ptr<ImageResource>
1> ]
1> and
1> [
1> T=std::shared_ptr<AnimationResource>
1> ]
1> d:\programming\github projects\boss\boss\new\resourcemanager\resourcemanager.h(21) : see reference to function template instantiation 'BaseMapManager<T> &ResourceManager::get_map<std::shared_ptr<_Ty>>(void)' being compiled
1> with
1> [
1> T=std::shared_ptr<AnimationResource>,
1> _Ty=AnimationResource
1> ]
1> d:\programming\github projects\boss\boss\new\main.cpp(17) : see reference to function template instantiation 'void ResourceManager::load_resource<AnimationResource>(std::string,std::string)' being compiled
1>d:\programming\github projects\boss\boss\new\resourcemanager\resourcemanager.h(44): error C2440: 'return' : cannot convert from 'BaseMapManager<T>' to 'BaseMapManager<T> &'
1> with
1> [
1> T=std::shared_ptr<FontResource>
1> ]
1> and
1> [
1> T=std::shared_ptr<AnimationResource>
1> ]
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
目标是 - 使用模板函数提供对不同类型数据的访问,例如
manager->get_resource<AnimationResource>(itsName)
//get it 或
manager->load_resource<FontResource>(itsPath, itsName)
//add it
有没有更好的方法来处理这个问题?谢谢!