Re: [BUG?] Segmentation fault in lua_error() - easily reproducible
- From: Sean Conner <sean@xxxxxxxxxx>
- To: luajit@xxxxxxxxxxxxx
- Date: Sun, 14 Dec 2014 20:56:56 -0500
It was thus said that the Great imzyxwvu@xxxxxxxxx once stated:
> > 在 2014年12月14日，17:38，"Sergei Zhirikov" <dmarc-noreply@xxxxxxxxxxxxx>
> > (Redacted sender "sfzhi@xxxxxxxxx" for DMARC) 写道：
> >
> > Back to the point, which is my original email. Please kindly ignore
> > everything written after that as irrelevant. However, to avoid any
> > further meaningless arguments, here is a modified version of the example
> > code, which does not change anything with respect to the segmentation
> > fault:
> >
> > #include <lua.h>
> > #include <lualib.h>
> > #include <lauxlib.h>
> > #include <stdio.h>
> >
> > int main()
> > {
> > lua_State *lua = luaL_newstate();
> > luaL_openlibs(lua);
> > lua_State *lua2 = lua_newthread(lua);
> > luaL_loadstring(lua2, "coroutine.yield()");
> > int res = lua_resume(lua2, 0);
> > printf("res=%d, top=%d\n", res, lua_gettop(lua2));
> > lua_pushliteral(lua2, "test");
> > lua_error(lua2);
> > return 0;
> > }
>
> Now there is already a main thread lua and a slave thread lua2. The core
> of this problem is trying to throw an error on an yielded thread lua2,
> whose behavior is *undefined*. No code can be executed on an yielded
> thread, so how can there be an error thrown on it? Did you try throwing an
> error on lua(which is not yielded) instead of lua2?
>
> There is no way to throw an error on an yielded thread with pure Lua
> code. At least I didn't find one.
Sergei is not trying to do that. What he is trying to report is a
condition that PUC Lua 5.1 handles:
[spc]lucy:/tmp>gcc -o s s.c -llua -lm -ldl
[spc]lucy:/tmp>./s
res=1, top=0
PANIC: unprotected error in call to Lua API (test)
That LuaJIT does not:
[spc]lucy:/tmp>gcc -o s s.c -lluajit-5.1 -lm -ldl
[spc]lucy:/tmp>./s
res=1, top=0
Segmentation fault (core dumped)
-spc
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