Yes, ou can use `dynamic-require` with a limited code inspector like
this:
(parameterize ([current-code-inspector (make-inspector)])
(dynamic-require 'untrusted-foo 'foo-provided-name))
At Fri, 22 Oct 2021 12:42:58 -0700 (PDT), "kalime...@gmail.com" wrote:
> Thank you!
>
> Is it possible
Thank you!
Is it possible to safely load untrusted module with dynamic-require?
пятница, 22 октября 2021 г. в 22:59:57 UTC+5, Robby Findler:
> On Fri, Oct 22, 2021 at 12:43 PM Matthew Flatt wrote:
>
>> At Thu, 21 Oct 2021 07:37:12 -0700 (PDT), "kalime...@gmail.com" wrote:
>> > I've read about
On Fri, Oct 22, 2021 at 12:43 PM Matthew Flatt wrote:
> At Thu, 21 Oct 2021 07:37:12 -0700 (PDT), "kalime...@gmail.com" wrote:
> > I've read about protect-out and current-code-inspector, but I still
> cannot
> > understand, how to require a module and forbid it to run protected
> modules.
> >
>
At Thu, 21 Oct 2021 07:37:12 -0700 (PDT), "kalime...@gmail.com" wrote:
> I've read about protect-out and current-code-inspector, but I still cannot
> understand, how to require a module and forbid it to run protected modules.
>
> Something like (require untrusted-foo) (foo-proc) but to forbid
I'd be interested to know this as well. It sounds like something that
isn't possible in Racket, since it's essentially specifying how a module
can do its job and that requires a level of introspection that I think is
excluded by design.
On Thu, Oct 21, 2021 at 10:37 AM kalime...@gmail.com
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