使用可變參數模板從Lua函數回調（C api）提取參數

Question

我正在嘗試編寫一個類，該類提取從Lua函數調用（Lua腳本調用已註冊的C函數）傳遞的參數，並將它們傳遞給已註冊方法（IDelegate在我的代碼片段），所以我可以執行它並返回值。

假設我寄存器從遊戲鍵盤類下面的方法： long int GameBoard::testFunct(long int);如「GB：testFunction」的Lua下用下面的代碼：

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 & MethodDelegatePtr用於註冊方法，函數和函數，所以我可以稍後給他們打電話）

然後LuaCall<I32,GameBoard,I32>::LuaCallback將被稱爲（以Lua堆棧爲參數）當我將在Lua腳本中寫入GB:testFunction(17);，然後註冊的方法將被觸發並返回等待的值。

它的工作原理如果我註冊並調用一個沒有任何參數的方法。 但如果等待爲long int GameBoard::testFunct(long int);任何參數做的話，我已經得到了以下錯誤......

In static member function 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]’:

instantiated from ‘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]’

instantiated from ‘static int LuaCall::LuaCallback(lua_State*) [with C = GameBoard, Args = {long int}, lua_State = lua_State]’

error: no match for call to ‘(MethodDelegatePtr) (std::basic_string&)’

note: no known conversion for argument 1 from ‘std::basic_string’ to ‘long int&’

---

In static member function ‘static Tr tUnpackLuaArgs<0u>::unpack(IDelegate*, lua_State*, ArgsValue ...) [with C = GameBoard, Tr = int, Args = {long int}, ArgsValue = {bool}, lua_State = lua_State]’:

instantiated from ‘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]’

instantiated from ‘static int LuaCall::LuaCallback(lua_State*) [with C = GameBoard, Args = {long int}, lua_State = lua_State]’

error: no match for call to ‘(MethodDelegatePtr) (bool&)’

note: no known conversion for argument 1 from ‘bool’ to ‘long int&’

我找不到爲什麼ArgsValue試圖通過一個std::basic_string<char>或當我註冊的方法bool等待long int ...它應該通過long int。

這裏是我寫的類來提取來自Lua腳本函數調用的參數。

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...); 
    } 
}; 

這裏是我如何使用它：

// 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; 
    } 
}; 

事實上，如果我刪除LUA_TSTRING和LUA_TBOOLEAN從的if/else在unpack功能，它編譯和工作正常。

Answer 1

就這樣，如果您的代碼包含例如tUnpackLuaArgs<1>::unpack，那麼unpack的主體將熱切地實例化全部的可能性（因爲它們都可能在運行時取決於Lua棧上參數的類型）。

這包括例如tUnpackLuaArgs<0>::apply<C, Tr, Args..., std::string>，這意味着當Args是例如long int由於std::string無法轉換爲long int const&，因此operator()的IDelegate預期。

您並不需要所有這些可能性：您已經知道您的期望（例如您的案例中的long int）。當您想要long int時提取std::string是沒有用的。所以相反，你應該嘗試從Lua堆棧中提取你期望的內容（如果轉換不起作用，可能會出錯，直到運行時纔會知道）。