Re: "Require" in a sandboxed environment

[Nick Gammon <nick@...>]

On 02/08/2007, at 5:48 PM, Tom Miles wrote:

> code from environment 1:
>
> utils = require "Utilities"
>
> utils.g_var = 1
> print(utils.g_var)
>
> -----------------------------------------------------
>
> code from environment 2:
>
> utils = require "Utilities"
>
> print(utils.g_var)

> In the lua code example given, env 1 and env 2 would both inherit a  
> common host environment, which itself is a child of the main state.  
> i.e.
>
> main->host->env 1
>                      ->env 2


OK, here is another guess. :)

env1 and env2 both have their own states with a metatable entry for  
__index for Lua globals. However under that setup "utils" is a local  
variable to both env1 and env2. That is, they both have their own  
"utils" variable. Thus you are not sharing "utils", and an attempt to  
print(utils.g_var) will not reference the other "utils".

What *may* work is to do this at the global state (that is, not in a  
sub-environment):

utils = require "Utilities"

This establishes utils at global scope.

- Nick