Re: how does toLua protect against double delete?

[Ariel Manzur <listas@...>]

From what I've seen, tolua only sends the object to the clone luastate 
when the function is returning the object by value instead of by pointer or 
reference, so that the value only survives the scope of the lua reference 
(if any) that receives it. So just like in C++, you can't (or shouldn't) 
delete an object you got from a function that returned it. Example:

class Vector3D {
        float x, y, z;
};
Vector3D position;

Vector3D get_position() { return position;};
Vector3D *get_new_vector() { return new Vector3D; };

In that case, toulua will only push the value from 'get_position' to the 
clone luastate, so that the copy of 'position' you got from the function 
gets deleted with the reference that contains it on lua. The value from 
'get_new_vector' is just a pointer, so you decide who's responsible to 
delete it; the only thing that dies with that lua reference is the value of 
the pointer (just like in C++).

Of course, I might be wrong (please correct me).

Hope that helps.. bye..

Ariel.

At 10:57 08/03/2003 -0800, you wrote:
> If you push a C++ object into Lua via toLua, tolua will use tolua_doclone()
> to make sure that it gets deleted when Lua loses all references to it. this
> is very cool. however, I'm curious what happens if the Lua script author
> does and explicit delete() on the object before undoclone() is called - is
> there some way to prevent this from happening?

Ariel.
http://Anime.com.ar
snm