So this seems the good way to go.
However, of course, this does not work if the return type is compatible with
Void:
(: f ((U String Integer) -> Any))
(define (f x)
(cond
[(string? x) (string=? x "hi")]
[(exact-nonnegative-integer? x) (= x 7)]))
typechecks.
I wonder if there a trick to make this work in all cases.
Probably not with standard `cond', because as Ian suggested, this would require
to specify an else clause that returns something which is not a subtype of Any.
I guess you could include a type Invalid, which is NOT a subtype of Any. That
would make it possible to provide a special `cond' form for which the proposed
pattern would work (the else clause would be typed as Invalid). Would something
like that work?
I guess your answer will be that Any is the only type with which Void is
compatible, so it is not "so much of a problem in practice".
Thanks,
-- Éric
On Oct 20, 2011, at 5:03 PM, Sam Tobin-Hochstadt wrote:
> On Thu, Oct 20, 2011 at 3:08 PM, J. Ian Johnson <i...@ccs.neu.edu> wrote:
>> Does typed racket fail to typecheck because cond will return void on
>> negative numbers or something more related to exhaustive checking regardless
>> of return value?
>
> The former. Since there's no pattern matching in "core" Racket, it's
> not even clear what exhaustiveness would mean in that case.
>
>> -Ian
>> ----- Original Message -----
>> From: Sam Tobin-Hochstadt <sa...@ccs.neu.edu>
>> To: Eric Tanter <etan...@dcc.uchile.cl>
>> Cc: Racket Users <users@racket-lang.org>
>> Sent: Thu, 20 Oct 2011 14:39:19 -0400 (EDT)
>> Subject: Re: [racket] [typed racket] type case
>>
>> There are basically two options here:
>>
>> 1. If you just use `cond' with no `else', you get a type error when
>> your cases are not exhaustive. For example:
>>
>> (: f ((U String Integer) -> Boolean))
>> (define (f x)
>> [(string? x) (string=? x "hi")]
>> [(exact-nonnegative-integer? x) (= x 7)])
>>
>> This program will fail to typecheck.
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