CODEWITHSUNDEEP
c++lua
user1241335
user1241335

Reputation:

binding lua in static library segfault

I'm trying to bind a C++ (functional) lua extension as a static library to a host C++ program.
I made a minimalistic program to see if it's going to work properly. I currently have this code:

main.cpp :

#include "util.h"

int main()
{
    lua_State* L;
    startLua(L);//after the call to this is gets the segfault
    luaL_dofile(L,"test.lua");
}

util.h :

#ifndef _UTIL_FILE_INCLUDED_
#define _UTIL_FILE_INCLUDED_

extern "C" {
#include <lua.h>
#include <lualib.h>
#include <lauxlib.h>
}


//functions
int startLua(lua_State* start);

#endif

util.cpp :

#include "util.h"

int startLua(lua_State* start)
{
    //program never gets here
    start = lua_open();
    luaL_openlibs(start);
    return 0;
}

Makefile :

CC=g++
CFLAGS=-c -Wall -fpermissive 
LFLAGS=-llua


all: util.a main.o 
    $(CC) $(LFLAGS) main.o util.a -o main
util.a: util.o util.h
    ar -cvq util.a util.o
util.o: util.cpp util.h
    $(CC) $(CFLAGS) util.cpp
main.o: main.cpp util.h
    $(CC) $(CFLAGS) main.cpp
clean:
    rm *.o *.a

I also tried undefining L from the header and defining it purely inside of main(), with same result.

the only warning I get during compile is that "L is defined but not used" or "L is used uninitialized".

at runtime it starts by giving me out a segfault right away.

I don't exactly work with libraries that often, so it may be a fault there, in that case this should be a simple question.

In case it matters, I use gcc 4.7.2, lua 5.1.5 and Arch Linux (updated).

EDIT: According to GDB, the problem is here: Program received signal SIGSEGV, Segmentation fault. 0x00007ffff7bb6390 in lua_gettop () from /usr/lib/liblua.so.5.1

I never had problems with lua, is it possible that compiling it into a static library is a mistake?

Upvotes: 0

Views: 637

Answers (1)

Nicol Bolas
Nicol Bolas

Reputation: 473232

lua_State* L;
startLua(L);//after the call to this is gets the segfault

This is a basic C problem here. You cannot change the value of parameters in C when they are passed by value.

startLua sets the value of its parameter. But that doesn't affect L at all. This isn't C++ (and even if it were, you didn't pass it by reference). If you want to affect L, then you must pass a pointer to L itself. As in &L, and startLua has to take a lua_State**. And to set the value, it must dereference the pointer (*start = lua_open()).

Or even better, just do this:

lua_State *L = startLua();

And have startLua return the Lua state.

Upvotes: 2

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