Question

我正在尝试将纹理从Lua脚本加载到我的C ++游戏引擎中。

引擎使用名为＆＃34; ResourceHolder＆＃34;的类。枚举类型来自名为＆＃34; ResourceIdenifiers＆＃34;。

的类

我的游戏场景为Textures＆amp ;;创建了自己的ResourceHolder。字体（以及我需要的任何其他内容）。所以我有Textures :: ID（枚举类型）和Fonts :: ID。

的命名空间

所以我只需创建一个TextureHolder对象＆m ;;纹理＆＃39;

TextureHolder                       mTextures;

然后我简单地用一条单线加载纹理，如下所示：

mTextures.load(Textures::Airplane2, "../GFX/Airplane2.png");

问题在于我不能在Lua中使用这些枚举类型，尽管我计划在我的lua.script文件中使用这样的类型：

allTextures
{

--Airplanes
["Airplane1"]       = "../GFX/Airplane1.png",
["Airplane2"]       = "../GFX/Airplane2.png",

--Or something like this instead
["Textures::Airplane3"]         = "../GFX/Airplane3.png"

}

允许Lua脚本处理这些枚举类型的最简单方法是什么？

以下是ResourceIdentifier和ResourceHolder的类。

ResourceIdentifier.h

#ifndef RESOURCEIDENTIFIERS_H
#define RESOURCEIDENTIFIERS_H


// Forward declaration of SFML classes
namespace sf
{
class Texture;
class Font;
}

namespace Textures
{
enum ID
{
    //Airplanes
    Airplane1,
    Airplane2,
    Airplane3,
    Background1,
    Background2,
};
}

namespace Fonts
{
enum ID
{
    Main,
};
}

// Forward declaration and a few type definitions
template <typename Resource, typename Identifier>
class ResourceHolder;

typedef ResourceHolder<sf::Texture, Textures::ID> TextureHolder;
typedef ResourceHolder<sf::Font, Fonts::ID>         FontHolder;

#endif // RESOURCEIDENTIFIERS_H

ResourceHolder.h（不太相关）

#ifndef RESOURCEHOLDER_H
#define RESOURCEHOLDER_H

#include <map>
#include <string>
#include <memory>
#include <stdexcept>
#include <cassert>

#include <SFML/Graphics/Image.hpp>

template <typename Resource, typename Identifier>

//This class stores Identifier so they can be accessed.
class ResourceHolder
{
public:
    //This creates loads the texture from the filename, gives it an ID, and stores it in the std::map container mTextureMap.
    void load(Identifier id, const std::string& filename);

    void loadImage(Identifier id, const sf::Image& image);

    template <typename Parameter>
    void load(Identifier id, const std::string& filename, const Parameter& secondParam);

    //This gets the texture from the std::map container, so it can be used. It gets the Resource based on the texture's ID (name).
    Resource& get(Identifier id);
    const Resource& get(Identifier id) const;
    //^SFML book - Chapter 2 - "Accessing the Identifier" ??? For when you dont want to allow editing of the Texture???


private:
    //A map stores all of the Identifier. The std::map< (1 parameter) 'Name of Resource', (2 parameter) a unique pointer of the Resource).
    std::map<Identifier, std::unique_ptr<Resource> > mResourceMap;

};

#include "ResourceHolder.inl"

#endif // RESOURCEHOLDER_H

Answer 1

这是最简单的方法是实际上只使用Lua中基本上是新手的枚举类型，通过将枚举类型从C ++复制/粘贴到Lua（并删除逗号），然后将纹理加载到图像按名称和路径。

我使用的是“LuaPlus”而不是LuaBridge。虽然LuaPlus是要安装的PITA，但在正确添加到项目中时它更优越（更易于使用和理解）。

以下是我列举的类型：

namespace Textures
{
    enum ID
    {
        //Airplanes
        Airplane1 = 1,
        Airplane2 = 2,
        Airplane3 = 3,
        //Backgrounds
        Background1 = 100,
        Background2 = 101,
    };
}

我的lua脚本最终看起来像这样：

--Copy/Paste the Enumerated Types here, deleting the "," commas.
Airplane1 = 1
Airplane2 = 2
Airplane3 = 3
Background1 = 100
Background2 = 101

--All Textures are registered here.
allTextures =
{

--Airplanes
[Airplane1]         = "../GFX/Airplane1.png",
[Airplane2]         = "../GFX/Airplane2.png",
--Backgrounds
[Background]        = "../GFX/Background.png",

}

然后在C ++中调用函数

//LOAD GAME TEXTURES HERE
void Scene::loadTextures()
{
LuaState* pLuaState = LuaState::Create();
pLuaState->DoFile("../GFX/test.lua");
LuaObject table = pLuaState->GetGlobals()["allTextures"];

for(LuaTableIterator it(table); it; it.Next())
{
    int key = it.GetKey().GetInteger();
    const char* value = it.GetValue().GetString();
    //std::cout<<"Key: "<<key<<", Value: "<<value<<std::endl;
    mTextures.load(static_cast<Textures::ID>(key), value);
}

}

将枚举类型从Lua传递给C ++的最简单方法是什么？

1 个答案: