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- Subject: luaopen_io() gives panic...
- From: L-28C <kixdemp@...>
- Date: Fri, 30 Mar 2007 21:41:37 -0400
<-- \code\ -->
#include <stdio.h>
#include <lua.h>
#include <lauxlib.h>
#include <lualib.h>
#include "mtx.luc.h"
int main(int argc, char *argv[])
{
int a;
// Open Lua and its libraries
lua_State *L = lua_open();
luaopen_io(L);
// Run!
if (luaL_loadbuffer(L,(const char*)B1,sizeof(B1),"mtx.luc")==0)
{
lua_pushstring(L, "arg");
lua_createtable(L, argc, 0);
for (a = 0; a < argc; a++)
lua_pushstring(L, argv[a]);
lua_settable(L, LUA_GLOBALSINDEX);
if (lua_pcall(L, 1, 0, 0) != 0)
puts(lua_tostring(L,-1));
}
// Close Lua and exit program
lua_close(L);
return 0;
}
<-- /code/ -->
It gives me error:
PANIC: unprotected error in call to Lua API (no calling environment)
However, if I remove the luaopen_io() call, I get:
?:0: attempt to index global 'io' (a nil value)
Any ideas why? I found an Internet post, and it got solved, but the guy
didn't say how. >:-@
Thanks in advance!