Casting between void * and a pointer to member function

Question

I\'m currently using GCC 4.4, and I\'m having quite the headache casting between void* and a pointer to member function. I\'m trying to write an easy-to-use li

Answer 1

It is possible to convert pointer to member functions and attributes using unions:

// helper union to cast pointer to member
template<typename classT, typename memberT>
union u_ptm_cast {
    memberT classT::*pmember;
    void *pvoid;
};

To convert, put the source value into one member, and pull the target value out of the other.

While this method is practical, I have no idea if it's going to work in every case.

Answer 2

Here, just change the parameters of the function void_cast for it to fit your needs:

template<typename T, typename R>
void* void_cast(R(T::*f)())
{
    union
    {
        R(T::*pf)();
        void* p;
    };
    pf = f;
    return p;
}

example use:

auto pvoid = void_cast(&Foo::foo);

Answer 3

Unlike the address of a nonstatic member function, which is a pointer-to-member type with a complicated representation, the address of a static member function is usually a just a machine address, compatible with a conversion to void *.

If you need to bind a C++ non-static member function to a C or C-like callback mechanism based on void *, what you can try to do is write a static wrapper instead.

The wrapper can take a pointer to an instance as an argument, and pass control to the nonstatic member function:

void myclass::static_fun(myclass *instance, int arg)
{
   instance->nonstatic_fun(arg);
}

Answer 4

You cannot cast a pointer-to-member to void * or to any other "regular" pointer type. Pointers-to-members are not addresses the way regular pointers are. What you most likely will need to do is wrap your member function in a regular function. The C++ FAQ Lite explains this in some detail. The main issue is that the data needed to implement a pointer-to-member is not just an address, and in fact varies tremendously based on the compiler implementation.

I presume you have control over what the user data lua_touserdata is returning. It can't be a pointer-to-member since there isn't a legal way to get this information back out. But you do have some other choices:

• The simplest choice is probably to wrap your member function in a free function and return that. That free function should take the object as its first argument. See the code sample below.

• Use a technique similar to that of Boost.Bind's mem_fun to return a function object, which you can template on appropriately. I don't see that this is easier, but it would let you associate the more state with the function return if you needed to.

Here's a rewrite of your function using the first way:

template <class T>
int call_int_function(lua_State *L) 
{
    void (*method)(T*, int, int) = reinterpret_cast<void (*)(T*, int, int)>(lua_touserdata(L, lua_upvalueindex(1)));
    T *obj = reinterpret_cast<T *>(lua_touserdata(L, 1));

   method(obj, lua_tointeger(L, 2), lua_tointeger(L, 3));
   return 0;
}

Answer 5

As a workaround given the restrictions of casting a pointer-to-member-function to void* you could wrap the function pointer in a small heap-allocated struct and put a pointer to that struct in your Lua user data:

template <typename T>
struct LuaUserData {
    typename void (T::*MemberProc)(int, int);

    explicit LuaUserData(MemberProc proc) :
        mProc(proc)
    { }

    MemberProc mProc;
};

LuaObject<Foo> lobj = registerObject(L, "foo", fooObject);
LuaUserData<Foo>* lobj_data = new LuaUserData<Foo>(&Foo::bar);

lobj.addField(L, "bar", lobj_data);

// ...

template <class T>
int call_int_function(lua_State *L) 
{
    typedef LuaUserData<T>                       LuaUserDataType;
    typedef typename LuaUserDataType::MemberProc ProcType;

    // this next line is problematic
    LuaUserDataType* data =
        reinterpret_cast<LuaUserDataType*>(lua_touserdata(L, lua_upvalueindex(1)));
    T *obj = reinterpret_cast<T *>(lua_touserdata(L, 1));

    (obj->*(data.mMemberProc))(lua_tointeger(L, 2), lua_tointeger(L, 3));
    return 0;
}

I'm not savvy with Lua so I have likely overlooked something in the above example. Keep in mind, too, if you go this route you'll have to manage the LuaUserData's allocation.