Re: Bin2c does nothing in my app?

[Robert Raschke <rrlua@...>]

L-28C writes:
> Okay. I bin2c'd a luac'd file. I included the bin2c output in my C++ 
> program, it gave syntax errors, an opening brace at the top and a 
> function call at the bottom. I removed the brace and cut/pasted the func 
> call to my C++ program. Here:
> 
> -- CODE/ --
> #include <stdio.h>
> #include <lua.h>
> #include <lauxlib.h>
> #include <lualib.h>
> #include "mtx.luc.h" // bin2c file
> 
> int main()
> {
> 	lua_State *L = lua_open();
> 	if (luaL_loadbuffer(L,(const char*)B1,sizeof(B1),"mtx.luc")==0) 
> lua_pcall(L, 0, 0, 0);
> 	lua_close(L);
> 	return 0;
> }
> -- /CODE --
> 
> I run the program but it gives no output. I'm on VC++ 2005 and it 
> compiles just fine.

Hmm, I use bin2c under Lua 5.0.3 and my C code looks roughly like this:

static int
open_mtx(lua_State *L)
{
#include "mtx.luc.h"
return 0;
}

int
main()
{
	lua_State *L = lua_open();
	open_mtx(L);
	// ...
}

As far as I understand, bin2c creates a bit of code that can be
executed.  So, it needs to be included somewhere in your C++ code
where you can then call it.  That's why the #include is inside the
function.  When you execute the code, then the compiled lua is run.  I
guess you could simplify this even further (I don't because of lots of
other stuff in my code):

int main()
main()
{
	lua_State *L = lua_open();
#	include "mtx.luc.h"
	// ...
}

The bin2c code references L, so you'll always have to make the
provision for L to be an available lua_State in the scope where you
include the bin2c code.

Robby

--
r dot raschke at tombob dot com