Hello,
It is impossible to pass object types that are modelled in lua
with built-in types by reference. In Lua, there exists no
attribute to types like reference or value type. A type is either
a reference type (tables), or a value type (built-in types like
number).
What you can do however is return the new value in the lua-side. A
policy could be established, that transforms reference integral
types to a list of parsed return types, so it is transparent on
the c++ side. That would place hard constraints on the lua side
though.
Kind regards,
Michael
Am 03.08.2014 21:09, schrieb 1 1:
When programming on C++, you can do the following:
void byReference(int &y)
{
y = 5;
}
int main()
{
int x = 2; // x = 2
byReference(x); // x = 5
}
How to do the same using luabind?
Luabind docs says:
If you want to pass a parameter as a reference, you have
to wrap it with the Boost.Ref.
Like this:
int ret = call_function(L, "fun", boost::ref(val));
But when I'm trying to do this:
#include <iostream>
#include <conio.h>
extern "C"
{
#include "lua.h"
#include "lualib.h"
#include "lauxlib.h"
}
#include <luabind\luabind.hpp>
using namespace std;
using namespace luabind;
int main()
{
lua_State *myLuaState = luaL_newstate();
open(myLuaState);
int x = 2;
do
{
luaL_dofile(myLuaState, "script.lua");
cout<<"x before = "<< x <<endl;
try
{
call_function<void>(myLuaState, "test", boost::ref(x));
}
catch(const std::exception &TheError)
{
cerr << TheError.what() << endl;
}
cout<<"x after = "<< x <<endl;
} while(_getch() != 27);
lua_close(myLuaState);
}
script.lua:
function test(x)
x = 7
end
It gives me a "Trying to use unregistered class"
error.
So, how to pass value from C++ to lua function by
reference, so I can change it inside the script and it will
be changed in c++ program too? I'm using boost 1.55.0, lua
5.1, luabind 0.9.1
Thanks.
P.S:
I posted the same question on
StackOverflow,
but seems like nobody knows what to do.
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