I catched the trick: when the argument will not be modified, declaring it const does the job. But when it will be modified (as in my example for recv() ) to have in the middle of lua code a call to malloc() and then to free() is not a "good thing" to my taste.
What about local v = ffi.new( "char[?]", bufsize ) Is it ok? TIAPS: what if malloc() cannot allocate any memory? NULL from malloc comes as nil to lua code?
On 11/19/2014 04:49 PM, Florian Weimer wrote:
* marcelo nicolet:This answer was right: -Wl, -E did the trick. Now, being the function create_thread( char * ), and calling it as C.create_thread( lua_string ) it gives ./mainthread: second.lua:33: bad argument #1 to 'create_thread' (cannot convert 'string' to 'char *') What can I do? How can I convert the lua string to char*You need to call strdup (and free the pointer somewhere) if the create_thread function indeed modifies the argument string. Otherwise, it is sufficient to change the type of the argument to “const char *” (both in the Lua declaration and the C implementation). Lua strings must not be modified, and the C type system enforces this for once. (ffi.cast is the wrong way to tackle this.)
