Re: lua_State *L in a Lua/C function called via FFI
- From: Duncan Cross <duncan.cross@xxxxxxxxx>
- To: luajit@xxxxxxxxxxxxx
- Date: Wed, 31 Oct 2012 15:13:57 +0000
On Wed, Oct 31, 2012 at 2:33 PM, François Perrad
<francois.perrad@xxxxxxxx> wrote:
> I try to use some Lua/C functions which are included in the ffi.C namespace.
> More especially, I refactore a code which uses C malloc/free, with lua_Alloc.
>
> local ffi = require "ffi"
> ffi.cdef[[
> typedef struct lua_State lua_State;
> typedef void * (*lua_Alloc) (void *ud, void *ptr, size_t osize,
> size_t nsize);
> lua_Alloc lua_getallocf (lua_State *L, void **ud);
> ]]
> print(ffi.C.lua_getallocf) --> cdata<void *(*())()>: 0x080530d0
>
> local ud = ffi.new("void*")
> print(ud) --> cdata<void *>: NULL
> local alloc = ffi.C.lua_getallocf(L, ud)
> local buffer = alloc(ud, 0, 0, 4096)
>
> but I don't know how to retrieve the current lua_State which is the
> first parameter of all Lua/C functions.
>
> François
>
The FFI does not provide any way to access the current lua_State, by
design [1]. What you may be able to do is create a dummy coroutine and
use the thread object for L, but I haven't tried that myself.
By the way, assuming you do manage to get a valid L, there is a
separate issue with your example code: passing a "void*" value for the
'ud' parameter will not work the way you expect. You should instead do
something like this:
local out_ud = ffi.new("void*[1]")
local alloc = ffi.C.lua_getallocf(L, out_ud)
local buffer = alloc(out_ud[0], 0, 0, 4096)
[1] http://lua-users.org/lists/lua-l/2011-10/msg01075.html
-Duncan
Other related posts:
