[racket-dev] syntax-local-inferred-name, keyword functions, and syntax properties

Aha, I found the example:

   (let ([foo ((lambda (#:kk x) (λ (x) x)) #:k 0)]) (foo 5))

Has this error:

application: procedure does not expect an argument with given keyword
  procedure: foo
  given keyword: #:k
  arguments...:
   #:k 0


Note that if you make the keywords the same, `foo` is indeed bound to
a procedure, but _not_ to the procedure referred to in the error
message.

Sam

On Mon, Dec 9, 2013 at 4:29 PM, Sam Tobin-Hochstadt
<samth at cs.indiana.edu> wrote:
> I haven't found a way to make it happen yet.  But even so, it seems
> like the wrong name.
>
> Sam
>
> On Mon, Dec 9, 2013 at 4:16 PM, Robby Findler
> <robby at eecs.northwestern.edu> wrote:
>> Is it possible that that name can leak out in an error message?
>>
>> Robby
>>
>>
>>
>> On Mon, Dec 9, 2013 at 3:10 PM, Sam Tobin-Hochstadt <samth at cs.indiana.edu>
>> wrote:
>>>
>>> Currently, this program fails with a somewhat bizarre type error:
>>>
>>>     #lang typed/racket
>>>
>>>     (: foo ([#:k Any] -> Integer))
>>>     (define (foo #:k [s #f]) 0)
>>>
>>>     (let: ([i : Integer (foo #:k #t)]) i)
>>>
>>> The reason is that the expansion of keyword applications generates a
>>> name to use for the function (here `foo`), and it uses
>>> `syntax-local-infer-name` to get the name to use. Unfortunately, in
>>> this case, it produces `i`, an identifier which has an extra syntax
>>> property saying that `i` is an `Integer`.  Of course, `foo` isn't an
>>> integer, it's a function, and so we get a type error.
>>>
>>> I don't see why the inferred name is the right choice here -- there's
>>> no connection between `i` and the name of the function. I can just
>>> change this to use a fresh name, but I thought I'd ask first.
>>>
>>> Sam
>>> _________________________
>>>   Racket Developers list:
>>>   http://lists.racket-lang.org/dev
>>
>>