如何向Lua公开C ++联合 [英] How to expose C++ unions to Lua

问题描述

对于我正在研究的项目，我需要向Lua公开另一个库中的一些C ++类.不幸的是，该库中最重要的类之一具有许多联合和枚举( SFML 中的sf :: Event类)，以及

For a project I'm working on, I need to expose some C++ classes in another library to Lua. Unfortunately, one of the most important classes in this library has lots of Unions and Enums (the sf::Event class from SFML), and from a quick Google search I've discovered there is nothing on exposing C++ Unions to Lua. I don't mind if it's being exposed with the Lua/C API, with a library or with a binding generator, as long as it works. But, I'd prefer not to use a binding generator, because I want to be able to create an object in C++, then expose that instance of the object to Lua (unless that's possible with a binding generator)

推荐答案

要注册C/C ++函数，您首先需要使函数看起来像Lua提供的标准C函数模式:

For registering C/C++ functions you need to first make your function look like a standard C function pattern which Lua provides:

extern "C" int MyFunc(lua_State* L)
{
    int a = lua_tointeger(L, 1); // First argument
    int b = lua_tointeger(L, 2); // Second argument
    int result = a + b;

    lua_pushinteger(L, result);

    return 1; // Count of returned values
}

需要在Lua中注册的每个功能都应遵循此模式.返回类型int和lua_State* L的单个参数.并计算返回值. 然后，您需要在Lua的register表中注册它，以便可以将其公开到脚本的上下文中:

Every function that needs to be registered in Lua should follow this pattern. Return type of int , single parameter of lua_State* L. And count of returned values. Then, you need to register it in Lua's register table so you can expose it to your script's context:

lua_register(L, "MyFunc", MyFunc);

要注册简单变量，您可以这样编写:

For registering simple variables you can write this:

lua_pushinteger(L, 10);
lua_setglobal(L, "MyVar");

之后，您可以从Lua脚本中调用函数.请记住，在运行具有用于注册对象的特定Lua状态的脚本之前，应先注册所有对象.

After that, you're able to call your function from a Lua script. Keep in mind that you should register all of your objects before running any script with that specific Lua state that you've used to register them.

在Lua:

print(MyFunc(10, MyVar))

结果:

20

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