Help with boost bind/functions

问题

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?

回答1:

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::function is 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 of self. So it has a code pointer for the code, and some data - and the target API expects just the code pointer.

回答2:

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.

function<int(lua_State *)> f = boost::bind(&LuaEngine::testFunc, this)
registerFunction<LuaEngine>("testFunc", f);

And your function expects a void return type and that is needed to change to int too.

来源：https://stackoverflow.com/questions/1413744/help-with-boost-bind-functions