> On Tue, 19 Jan 2010 17:49:47 +0100
> ingmar wirths <
ingmania@googlemail.com> wrote:
>
>> > The idea is that if a Lua function can't see another Lua function,
>> > it can't call it, and it can't magic up a reference to it (unless
>> > you expose the debug interface, of course. :)
>>
>> Allright, but since these restrictions are all implemented in lua, how
>> can i enforce them, given that the user should be permitted to modify
>> his ai? From the point of my limited knowledge so far, i guess the
>> user could trivially escape from a sandbox, by just rewriting
>> everything. Am i missing something here?
>
> Because to /create/ the sandbox requires the calling of functions that
> must be written in C and are included by default. You simply don't put
> those functions in your sand box.
>
> Trivial example:
>
> function sandboxed_function()
> print "hello, world!"
> end
>
> setfenv(sandboxed_function, { print = print })
>
> sandboxed_function()
>
> Try altering the body of sandboxed_function as much as you like, the
> only thing it'll be able to call is print.
