[racket] Why this contract violation behaviour?

This is a special case just to make life more convenient: we know that
no matter what boundary 'amount' crosses, it is going to trigger a
violation and it will blame that module, so we don't wait.

Robby

On Fri, Nov 16, 2012 at 8:24 PM, Harry Spier <vasishtha.spier at gmail.com> wrote:
> Thanks Matthias.
>
> Why does the first case which doesn't cross a module boundary cause a
> contract violation.
>
> #lang racket
> (provide/contract (amount positive?))
> (define amount -7)
>
>
>
> On Fri, Nov 16, 2012 at 8:24 PM, Matthias Felleisen <matthias at ccs.neu.edu>
> wrote:
>>
>>
>> Contracts really are like business contracts. They are between two
>> distinct parties; they are not by a party on itself.
>>
>> On occasion, you really do want to wish to break the contract inside a
>> module. This is well-known in the OO world, and it is often called the
>> re-entrance problem. Say you want a balanced tree object. Imagine it has a
>> method with a contract that requires 'balanced'ness but is also called
>> internally. Perhaps you don't need/wish to balance for internal calls. So
>> you do -- and you don't get caught.
>>
>> Contracts impose a run-time cost. To make the cost reasonable, we trust
>> the programmer and we let the programmer know that we trust him.
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> On Nov 16, 2012, at 7:52 PM, Galler wrote:
>>
>> >>
>> >> Even if you use define/contract, you may violate the contract on
>> >> recursive
>> > calls. This is intentional.
>> >>
>> >
>> > Thanks.
>> >
>> > I hadn't run into that behavior before and wasn't aware.
>> >
>> > Could you possibly elaborate on the thinking wrt this design choice?
>> >
>> > (define/contract (my-natural-number n)
>> >  (-> exact-positive-integer? any/c)
>> >  (print n) (newline)
>> >  (my-natural-number (sub1 n)))
>> >
>> > (my-natural-number 1)
>> > 1
>> > 0
>> > -1
>> > -2
>> > -3
>> > -4
>> >
>> >
>> >
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