我正在搞乱可变参数模板，并试图制定一个luabinding计划。我知道它们存在但只是试图搞乱新功能。我现在正在进行VS2013试验。我创建了一个函子类，它是我需要从C ++端调用lua函数的信息的包装器;但是，我不能为我的生活得到我的模板功能，将变量推入堆栈以解压缩和工作。我已经尝试了网上的每个例子我都能找到并尝试各种方式提到如何解包函数模板的值。我显然错过了一些东西。我注意到的所有例子都是＆amp;＆amp;移动参考。这是一个要求吗？
/*
These functions push an item onto the lua stack
*/
template<typename T> inline void lua_push(lua_State* L,T&& t){ static_assert(false, "Unsupported Type! Cannot set to lua stack"); }
template<> inline void lua_push<lua_CFunction>(lua_State* L,lua_CFunction&& func){ assert(L != nullptr && func != nullptr); lua_pushcfunction(L, func); }
template<> inline void lua_push<double>(lua_State* L,double&& d){ assert(L != nullptr); lua_pushnumber(L, d); }
template<> inline void lua_push<int>(lua_State* L,int&& i){ assert(L != nullptr); lua_pushinteger(L, i); }
template<> inline void lua_push<bool>(lua_State* L,bool&& b){ assert(L != nullptr); lua_pushboolean(L, b); }
template<> inline void lua_push<std::string>(lua_State* L,std::string&& s){ assert(L != nullptr); lua_pushlstring(L, s.c_str(), s.size()); }
template<> inline void lua_push<const char*>(lua_State* L,const char*&& s){ assert(L != nullptr); lua_pushstring(L, s); }
然后我想在这个课程中解压缩
template<typename Return,typename... Args> class LuaFunctor{};
/*
A Lua function that will return a single value.
*/
template<typename Return,typename... Args> class LuaFunctor<Return(Args...)>
{
private:
//The lua state the function exists on
lua_State* m_luaState;
//Name of the function to be called in lua
std::string m_FunctionName;
public:
//Return typedef
typedef Return return_type;
//The number of arguments the functor accepts
static const int arguments = sizeof...(Args);
//Constructors
inline LuaFunctor(lua_State* L,const std::string& name) : m_luaState(L), m_FunctionName(name) {}
inline LuaFunctor(lua_State* L,const char* name) : m_luaState(L), m_FunctionName(name) {}
//Function call overload that allows the functor to act like a function call of luascript
inline Return operator()(Args&&... args)
{
//Assert that the function name does exist and luaState is pointing to something hopefully meaningful
assert(m_luaState != nullptr && m_FunctionName.size() != 0);
//Set the function
lua_getglobal(m_luaState, m_FunctionName.c_str());
//Verify Lua function is pushed onto the stack
assert(lua_isfunction(m_luaState, -1));
//If arguments exist push them onto the stack
if (sizeof...(Args) != 0)
{
/*
How do I unpack this?????
I want to unpack this into multiple functions
One for each type of argument.
*/
lua_push(m_luaState, std::forward<Args>(args))...;
}
//Call the function that is in lua
int status = lua_pcall(m_luaState, sizeof...(Args), 1, 0);
/*
If there was an error calling the function throw an exception
TODO: parse the error using luas builtin decode of the error for now just pass it on
TODO: create lua_exception
*/
if (status != 0) throw std::exception("Error calling lua function");
//Return the value request by lua, error checking is built-in to the function to verify type
return lua_get<Return>(m_luaState);
}
};
答案 0 :(得分：4)
这是修复。递归函数调用。尝试过我正在阅读的各种黑客，但这很干净简单。
/*
These functions push an item onto the lua stack
*/
template<typename First, typename... Rest> inline void lua_push(lua_State* L,First first,Rest... rest)
{
lua_push(L, first);
lua_push(L, rest...);
}
template<typename T> inline void lua_push(lua_State* L, T t){ static_assert(false, "Invalid type attemptiing to be pushed onto lua stack!"); }
template<> inline void lua_push<lua_CFunction>(lua_State* L, lua_CFunction func){ assert(L != nullptr && func != nullptr); lua_pushcfunction(L, func); }
template<> inline void lua_push<double>(lua_State* L, double d){assert(L != nullptr); lua_pushnumber(L, d); }
template<> inline void lua_push<int>(lua_State* L, int i){ assert(L != nullptr); lua_pushinteger(L, i); }
template<> inline void lua_push<bool>(lua_State* L, bool b){ assert(L != nullptr); lua_pushboolean(L, b); }
template<> inline void lua_push<std::string>(lua_State* L, std::string s){assert(L != nullptr); lua_pushlstring(L, s.c_str(), s.size()); }
template<> inline void lua_push<const char*>(lua_State* L,const char* s){ assert(L != nullptr); lua_pushstring(L, s); }
现在代码完美无缺。
template<typename Return,typename... Args> class LuaFunctor<Return(Args...)>
{
private:
//The lua state the function exists on
lua_State* m_luaState;
//Name of the function to be called in lua
std::string m_FunctionName;
public:
//Return typedef
typedef Return return_type;
//The number of arguments the functor accepts
static const int arguments = sizeof...(Args);
//Constructors
inline LuaFunctor(lua_State* L,const std::string& name) : m_luaState(L), m_FunctionName(name) {}
inline LuaFunctor(lua_State* L,const char* name) : m_luaState(L), m_FunctionName(name) {}
//Function call overload that allows the functor to act like a function call of luascript
inline Return operator()(Args... args)
{
//Assert that the function name does exist and luaState is pointing to something hopefully meaningful
assert(m_luaState != nullptr && m_FunctionName.size() != 0);
//Set the function
lua_getglobal(m_luaState, m_FunctionName.c_str());
//Verify Lua function is pushed onto the stack
assert(lua_isfunction(m_luaState, -1));
//If arguments exist push them onto the stack
if (sizeof...(Args) != 0) lua_push(m_luaState, args...);
//Call the function that is in lua
int status = lua_pcall(m_luaState, sizeof...(Args), 1, 0);
/*
If there was an error calling the function throw an exception
TODO: parse the error using luas builtin decode of the error for now just pass it on
TODO: create lua_exception
*/
if (status != 0)
{
report_errors(status);
throw std::exception("Error calling lua function");
}
//Return the value request by lua, error checking is built-in to the function to verify type
return lua_get<Return>(m_luaState);
}
};
答案 1 :(得分：1)
最简单的方法是将lua_push更改为一堆覆盖。
让他们返回一些副本或可移动类型（bool会做，或struct nothing{};）。
然后：
auto sink[]={lua_push(m_luaState, std::forward<Args>(args))...};
将它们全部调用。