[racket-dev] A tricky chaperone puzzle

Ah, nope. That model doesn't include function chaperones!

Robby

On Thu, Jul 24, 2014 at 6:14 PM, Robby Findler
<robby at eecs.northwestern.edu> wrote:
> I also lean towards #2. What does the redex model say? Most of those
> pieces are in it, I think.
>
> Robby
>
> On Thu, Jul 24, 2014 at 3:25 PM, Matthew Flatt <mflatt at cs.utah.edu> wrote:
>> Nice example. Offhand, I think that #2 is right, but I'll have to look
>> at it more to be sure.
>>
>> At Thu, 24 Jul 2014 15:45:18 -0400, Sam Tobin-Hochstadt wrote:
>>> Consider the following module:
>>>
>>> (module m racket
>>>   (struct x [a])
>>>   (define v1 (x 'secret))
>>>   (define v2 (x 'public))
>>>   (provide v1 v2)
>>>   (provide/contract [x-a (-> x? (not/c 'secret))]))
>>>
>>> It appears that this ensures that you can't get 'secret. But, it turns
>>> out that I can write a function outside of `m` that behaves like `x-a`
>>> without the contract:
>>>
>>> (require (prefix-in m: 'm))
>>>
>>> (define (x-a v)
>>>   (define out #f)
>>>   (with-handlers ([void void])
>>>     (m:x-a (chaperone-struct v m:x-a (λ (s v) (set! out v) v))))
>>>   out)
>>>
>>> Now this works:
>>>
>>> (displayln (x-a m:v1)) ;; => 'secret
>>>
>>> The problem is that `m:x-a` is treated as a
>>> `struct-accessor-procedure?`, which is a capability for accessing the
>>> a field, even though it's a significantly restricted capability.
>>>
>>> There are a couple possible solutions I've thought of:
>>>
>>> 1. Require a non-chaperoned/impersonated accessor.
>>> 2. Actually use the chaperoned/impersonatored accessor to get the
>>> value out instead of the underlying accessor.
>>>
>>> 1 is a little less expressive. But note that 2 means that you have to
>>> only allow chaperoned procedures with `chaperone-struct`, and imposes
>>> significant complication on the runtime.
>>>
>>> I favor 1.
>>>
>>> Sam
>>>
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>>
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