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 <[email protected]> wrote:
>
>> At Thu, 21 Oct 2021 07:37:12 -0700 (PDT), "[email protected]" 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 
>> foo-proc to 
>> > use ffi/unsafe.
>>
>> If you use
>>
>>  (current-code-inspector (make-inspector))
>>  (require untrusted-foo)
>>
>>
> Just in case: I think Matthew as thinking of two subsequent REPL 
> interactions (or calls to eval or suchlike). If you put those two together 
> into a file in #lang racket, say, you won't be protected against 
> untrusted-foo.
>
> Robby
>  
>
>> and assuming that `untrusted-foo` hasn't been loaded earlier, then
>> `untrusted-foo` will not be able to use protected binding.
>>
>> That sequence will also disable the use of protected bindings by
>> anything that `untrusted-foo` depends on and that hasn't already been
>> loaded. So, if you want those dependencies to be able to use untrusted
>> things, you need to load the before `(current-code-inspector
>> (make-inspector))`.
>>
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>> .
>>
>

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