模板函数搞砸了

Question

所以即时通过我的ResourceManager类与使用C++的游戏进行一点问题。 所以我试图使我的常规addImage功能的模板功能，所以它会增加声音，但我有一些我不能真正处理的错误可以帮助我吗？ ：d

.HPP

#ifndef RESOURCE_MANAGER_HPP 
#define RESOURCE_MANAGER_HPP 

#include "Image.cpp" 
#include "SoundBuffer.cpp" 
#include <SFML/Graphics.hpp> 
#include <SFML/Audio.hpp> 

typedef std::map<std::string, sz::Image*> ImagesContainer; 
typedef std::map<std::string, sz::Image*>::iterator ImagesContainerIt; 
typedef std::map<std::string, sz::SoundBuffer*> SoundsContainer; 
typedef std::map<std::string, sz::SoundBuffer*>::iterator SoundsContainerIt; 
typedef std::map<std::string, sf::Music*> MusicContainer; 
typedef std::map<std::string, sf::Music*>::iterator MusicContainerIt; 

namespace sz 
{ 
     //  meanwhile this class is only for images, need to edit later for 
     //  it to be also able to load sounds, etc... 
     class ResourceManager{ 
       private: 
       ResourceManager() {}; 
       ResourceManager(ResourceManager const&) {}; 
       static ResourceManager *rm; 
       // add functions and variables here 
       ImagesContainer imagesContainer; 
       SoundsContainer soundsContainer; 
       MusicContainer musicContainer; 
       template <class type> 
       void AddNew(std::string imagePath); 

       public: 
       static ResourceManager *Instance(); 
       // add functions here 
       template <class type> 
       type *Get(std::string imagePath); 
     }; 
} 

#endif 

的.cpp

#include "ResourceManager.hpp" 
#include <typeinfo> 

namespace sz 
{ 
     ResourceManager *ResourceManager::rm = NULL; 

     ResourceManager *ResourceManager::Instance() 
     { 
      if (!rm) 
        rm = new ResourceManager; 

      return rm; 
     } 

     template <class type> 
     void ResourceManager::AddNew(std::string filePath) 
     { 
       type *item = new type(filePath); 
       if(typeid(type) == typeid(sz::Image)) 
         imagesContainer[filePath] = item; 
       else if(typeid(type) == typeid(sz::SoundBuffer)) 
         soundsContainer[filePath] = item; 
       else 
         return; 
     } 

     template <class type> 
     type *ResourceManager::Get(std::string filePath) 
     { 
       if(typeid(type) == typeid(sz::Image)) 
       { 
         ImagesContainerIt it = imagesContainer.find(filePath); 
         if(it == imagesContainer.end()) 
         { 
           AddNew<type>(filePath); 
         } 
         it = imagesContainer.find(filePath); 
         return it->second; 
       } 

       else if(typeid(type) == typeid(sz::SoundBuffer)) 
       { 
         SoundsContainerIt it = soundsContainer.find(filePath); 
         if(it == soundsContainer.end()) 
         { 
           AddNew<type>(filePath); 
         } 
         it = soundsContainer.find(filePath); 
         return it->second; 
       } 

       else 
         return NULL; 
     } 
} 

错误@ _ @

g++ -Wall -c "Sprite.cpp" (in directory: /home/gannash/Desktop/Open Heroes/Engine) 
In file included from Sprite.cpp:2:0: 
ResourceManager.cpp: In member function ‘type* sz::ResourceManager::Get(std::string) [with type = sz::Image, std::string = std::basic_string<char>]’: 
Sprite.cpp:10:65: instantiated from here 
ResourceManager.cpp:50:15: error: cannot convert ‘sz::SoundBuffer*’ to ‘sz::Image*’ in return 
ResourceManager.cpp: In member function ‘void sz::ResourceManager::AddNew(std::string) [with type = sz::Image, std::string = std::basic_string<char>]’: 
ResourceManager.cpp:36:5: instantiated from ‘type* sz::ResourceManager::Get(std::string) [with type = sz::Image, std::string = std::basic_string<char>]’ 
Sprite.cpp:10:65: instantiated from here 
ResourceManager.cpp:23:4: error: cannot convert ‘sz::Image*’ to ‘std::map<std::basic_string<char>, sz::SoundBuffer*>::mapped_type {aka sz::SoundBuffer*}’ in assignment 
ResourceManager.cpp: In member function ‘type* sz::ResourceManager::Get(std::string) [with type = sz::Image, std::string = std::basic_string<char>]’: 
ResourceManager.cpp:55:2: warning: control reaches end of non-void function [-Wreturn-type] 
Compilation failed. 

Answer 1

好吧，这是我的想法后，看看代码和错误消息。

你不能像这样的函数返回图像和声音。他们是个人类型。你必须指定要返回的类型。当你将if/else放在模板中时，它仍然会检查它是否可以返回所有这些类型，并且会失败。

您必须返回Image和Sound继承的BaseClass *（或任何您想要调用它的）。

Answer 2

的模板功能已为编译器完全定义为能够使用它们。这意味着你必须将这些函数的主体从源文件移动到头文件。

编辑：

你必须检查你如何使用模板类。例如参见此错误消息：

ResourceManager.cpp:50:15: error: cannot convert ‘sz::SoundBuffer*’ to ‘sz::Image*’ in return

除非sz::SoundBuffer继承sz::Image你有类型之间的不匹配。

Answer 3

“编译时如果”通过模板部分专业化完成。但在这里它甚至不需要：

template<typename type> struct Container { 
    static std::map<std::string, type*> container; 
}; 
std::map<std::string, sz::Image*>& ImagesContainer = Container<sz::Image>::container; 
// etc... 
template <class type> 
void ResourceManager::AddNew(std::string filePath) 
{ 
    type *item = new type(filePath); 
    Container<type>::container[filePath] = item; 
}