[plt-scheme] Contract changes a parameter-procedure to a procedure. Why?

Thanks for the parameter/c tip Carl.

Re baked in:

So far contracts have worked pretty much as I expected on a practical
level. This is the first time the implementation poked through and
surprised me.  I probably do need to be careful about my C/C++ baggage
which maybe primes me to not even expect there would be both
parameter? and parameter/c, or why.

On Fri, Apr 16, 2010 at 1:56 PM, Robby Findler
<robby at eecs.northwestern.edu> wrote:
> FWIW, I think we're getting a better understanding of what exactly
> needs to be baked in to the language to do a better job of making such
> contracts work, between things like struct properties and chaperones
> (but the contract library has not yet caught up with the latest
> goodies in mz0>
>
> Robby
>
> On Fri, Apr 16, 2010 at 12:53 PM, Matthias Felleisen
> <matthias at ccs.neu.edu> wrote:
>>
>> (The real issue is that real contracts can't be implemented as a library;
>> they have to be 'baked into' the core of the language. But this is not for
>> here or now.)
>>
>>
>> On Apr 16, 2010, at 1:45 PM, Carl Eastlund wrote:
>>
>>> On Fri, Apr 16, 2010 at 1:30 PM, Greg Hendershott
>>> <greghendershott at gmail.com> wrote:
>>>>
>>>> I had the bright (?) idea of specifying a contract for a parameter. But:
>>>>
>>>> #lang scheme
>>>>
>>>> (define p1
>>>>  (make-parameter #f))
>>>>
>>>> (parameter? p1) ; --> #t
>>>> p1 ; --> #<procedure:parameter-procedure>
>>>>
>>>> (define/contract p2
>>>>  (() (boolean?) . ->* . (or/c boolean? void?))
>>>>  (make-parameter #f))
>>>>
>>>> (parameter? p2) ; --> #f   Why?
>>>> p2 ; --> #<procedure>      Why?
>>>>
>>>> I discovered this when I made a contract for a procedure that requires
>>>> its argument to be a parameter. I can't use parameter? I'd have to
>>>> settle for procedure?.
>>>>
>>>> Is this because the contract is implemented as a wrapper procedure,
>>>> and the parameter? predicate only sees the contract wrapper procedure
>>>> not the parameter-procedure inside?
>>>
>>> That's exactly it.  Use parameter/c instead; it constructs a
>>> parameter-friendly wrapper.
>>>
>>> --Carl
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