帮助增强绑定/功能
我有这个函数签名，我必须匹配
typedef int (*lua_CFunction) (lua_State *L);//target sig
这是到目前为止我所拥有的：
//somewhere else...
...
registerFunction<LuaEngine>("testFunc", &LuaEngine::testFunc, this);
...
//0 arg callback
void funcCallback0(boost::function<void ()> func, lua_State *state)
{
func();
}
template<typename SelfType>
void registerFunction(const std::string &funcName, boost::function<void (SelfType*)> func, SelfType *self)
{
//funcToCall has to match lua_CFunction
boost::function<void (lua_State *)> funcToCall = boost::bind(&LuaEngine::funcCallback0, this,
boost::bind(func, self), _1);
lua_register(_luaState, funcName.c_str(), funcToCall);
}
但是，在 lua_register(_luaState...)，它仍然抱怨转换问题
错误 1 错误 C2664： 'lua_pushcclosure'：无法转换 参数2来自 'boost::function' 为 'lua_CFunction'
有人知道如何解决这个问题吗？
I have this function signature I have to match
typedef int (*lua_CFunction) (lua_State *L);//target sig
Here's what I have so far:
//somewhere else...
...
registerFunction<LuaEngine>("testFunc", &LuaEngine::testFunc, this);
...
//0 arg callback
void funcCallback0(boost::function<void ()> func, lua_State *state)
{
func();
}
template<typename SelfType>
void registerFunction(const std::string &funcName, boost::function<void (SelfType*)> func, SelfType *self)
{
//funcToCall has to match lua_CFunction
boost::function<void (lua_State *)> funcToCall = boost::bind(&LuaEngine::funcCallback0, this,
boost::bind(func, self), _1);
lua_register(_luaState, funcName.c_str(), funcToCall);
}
However, at lua_register(_luaState..., it's still complaining about conversion issues
Error 1 error C2664:
'lua_pushcclosure' : cannot convert
parameter 2 from
'boost::function' to
'lua_CFunction'
Anyone know how this can be solved?
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这不能直接解决。 Lua API 想要您提供一个简单的函数指针 - 这只是一个代码指针，没有其他任何东西。同时，boost::function 是一个函数对象，它不可能转换为普通函数指针，因为 - 粗略地说 - 它不仅捕获代码，还捕获状态。在您的示例中，捕获的状态是 self 的值。因此它有一个代码指针和一些数据——目标 API 只需要代码指针。
This cannot be solved directly. Lua API wants a plain function pointers from you - that's just a code pointer, and nothing else. Meanwhile,
boost::functionis a function object, and there's no way it could possibly be convertible to a plain function pointer, because - roughly speaking - it captures not just the code, but also the state. In your example, the captured state is the value ofself. So it has a code pointer for the code, and some data - and the target API expects just the code pointer.问题是编译器无法推断出模板参数，因为存在隐式转换。
您需要将函数指针存储到函数对象中。
并且您的函数需要一个 void 返回类型，并且也需要更改为 int 。
The problem is that the compiler cant deduce the template parameter because there is an implicit conversion.
You need to store the function pointer into a function object.
And your function expects a void return type and that is needed to change to int too.