我正在尝试编写一个类来提取从 Lua 函数调用（调用已注册 C 函数的 Lua 脚本）传递的参数并将它们传递给已注册的方法（我的代码片段中的 IDelegate），以便我可以执行它并返回值.
假设我从 GameBoard 类中注册了以下方法
：在 Lua 下注册long int GameBoard::testFunct(long int);为“GB:testFunction” ，代码如下：
luaL_newmetatable(mState, "GameBoard");
lua_pushstring(mState, "__index");
lua_pushvalue(mState, -2);
lua_settable(mState, -3);
lua_pushstring(mState,"testFunction");
hbt::IDelegate<I32,I32>* ideleg = new MethodDelegatePtr<GameBoard,I32,I32>( NULL, &GameBoard::testFunct); // will be deleted later
lua_pushlightuserdata (mState, (IDelegate<I32,I32>*)ideleg);
lua_pushcclosure(mState, LuaCall<I32,GameBoard,I32>::LuaCallback,1);
lua_settable(mState,-3);
（IDelegate和MethodDelegatePtr用于注册方法、函数和仿函数，以便我稍后调用它们）
然后当我在 Lua 脚本中写入时LuaCall<I32,GameBoard,I32>::LuaCallback将调用（以 Lua 堆栈作为参数）GB:testFunction(17);，然后将触发注册的方法并返回等待的值。
如果我注册并调用一个没有任何参数的方法，它就可以工作。但是如果有任何参数等待long int GameBoard::testFunct(long int);，那么我有以下错误......
在静态成员函数中 static Tr tUnpackLuaArgs<0u>::unpack(IDelegate*, lua_State*, ArgsValue ...) [with C = GameBoard, Tr = int, Args = {long int}, ArgsValue = {std::basic_string} , lua_State = lua_State]':
从 'static Tr tUnpackLuaArgs::unpack(IDelegate*, lua_State*, ArgsValue ...) [with C = GameBoard, Tr = int, Args = {long int}, ArgsValue = {}, unsigned int i = 1u, lua_State = lua_State]'</p>
从 'static int LuaCall::LuaCallback(lua_State*) [with C = GameBoard, Args = {long int}, lua_State = lua_State]'实例化</p>
错误：不匹配调用 '(MethodDelegatePtr) (std::basic_string&)'</p>
注意：没有已知的参数 1 从 'std::basic_string' 到 'long int&' 的转换</p>
在静态成员函数 'static Tr tUnpackLuaArgs<0u>::unpack(IDelegate*, lua_State*, ArgsValue ...) [with C = GameBoard, Tr = int, Args = {long int}, ArgsValue = {bool} , lua_State = lua_State]':
从 'static Tr tUnpackLuaArgs::unpack(IDelegate*, lua_State*, ArgsValue ...) [with C = GameBoard, Tr = int, Args = {long int}, ArgsValue = {}, unsigned int i = 1u, lua_State = lua_State]'</p>
从 'static int LuaCall::LuaCallback(lua_State*) [with C = GameBoard, Args = {long int}, lua_State = lua_State]'实例化</p>
错误：不匹配调用 '(MethodDelegatePtr) (bool&)'</p>
注意：没有已知的参数 1 从 'bool' 到 'long int&' 的转换</p>
当我注册一个等待 a 的方法时，我无法找到为什么 ArgsValue 尝试传递 astd::basic_string<char>或a ...它应该传递 a 。boollong intlong int
这是我编写的用于提取来自 Lua 脚本函数调用的参数的类。
template< unsigned int i >
class tUnpackLuaArgs
{
public:
template< class C, class Tr, class... Args, class... ArgsValue >
static Tr unpack( IDelegate<C,Tr,Args...>* ideleg, lua_State *L, ArgsValue... argsVal)
{
int t = lua_type(L, i+1);
if( t == LUA_TNUMBER)
{
I32 tmpUint = lua_tonumber(L, i+1);
return tUnpackLuaArgs< i-1 >::unpack( ideleg, L, argsVal..., tmpUint);
}
else if( t == LUA_TSTRING)
{
std::string tmpStr = lua_tostring(L, i+1);
return tUnpackLuaArgs< i-1 >::unpack( ideleg, L, argsVal..., tmpStr);
}
else if( t == LUA_TBOOLEAN)
{
bool tmpBool = lua_toboolean(L, i+1);
return tUnpackLuaArgs< i-1 >::unpack( ideleg, L, argsVal..., tmpBool);
}
//etc.
}
};
template<>
class tUnpackLuaArgs<0>
{
public:
template< class C, class Tr, class... Args, class... ArgsValue >
static Tr unpack( IDelegate<C,Tr,Args...>* ideleg, lua_State *L, ArgsValue... argsVal)
{
//-- Execute the registered method using the LUA arguments
//-- and returns the returned value
return (*ideleg)( argsVal... );
}
};
这是我使用它的方式：
// Specialized template returning an integer
template <class C, class... Args>
struct LuaCall<int, C, Args...>
{
static int LuaCallback(lua_State *L)
{
//-- retrieve method "IDelegate" from Lua stack etc.
//-- then call tUnpackLuaArgs with the arguments to push the returned value onto the lua stack
lua_pushnumber(L, tUnpackLuaArgs< sizeof...(Args) >::unpack(funcPtr,L));
return 1;
}
};
事实上，如果我从 if/else in函数中删除LUA_TSTRINGand the ，它确实可以编译并且工作正常。LUA_TBOOLEANunpack