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Re: case-lambda* question

From: Daniel Llorens
Subject: Re: case-lambda* question
Date: Mon, 19 Nov 2012 10:43:51 +0100

On Nov 15, 2012, at 02:22, Daniel Hartwig wrote:

> On 14 November 2012 18:20, Daniel Llorens <address@hidden> wrote:
>>> When the doc. states keyword arguments do not contribute to the
>>> success of a match, it refers only to keyword arguments in the
>>> case-lambda clause, not at the call site.  This makes sense, otherwise
>>> it would inhibit writing functions that detect keywords internally
>>> from their rest arguments.
>> Do you mean something like this?
>> (define* (f a #:key x) x)
>> (define (g . args) (apply f args))
>> (g 0) -> #f
>> (g 0 #:x #t) -> #t
>> i.e. g must accept being called with 3 'arguments' so that it can forward 
>> the keyword args.
> I was thinking of:
> (define f
>  (case-lambda*
>   ((a . rest)
>    (if (memq #:x rest) …
> where #:x is picked up from inside rest.
> Based on the other error, I'd say that any case-lambda* with keyword
> arguments is matched with a rest argument instead, i.e. “a #:key x” is
> treated as “a . rest”.

Ok, I think I see how 

 (define f (case-lambda*
             ((x #:optional y) 1)
             ((x #:key y)      2)
             ((x y #:key z)    3)))

 (f #:y 2)

works. x is #:y and y is 2.

However, this should go to (x y #:key z) and it doesn't:

 (f 1 2 #:z 3) -> Odd length of keyword argument list

while given

(define g (case-lambda*
                       ((a #:key x) 1)
                       ((a b c #:key x) 3)))

this should match (a b c #:key x) and it doesn't. 

(g 1 2 3) - > Invalid keyword

Or maybe we should say that (g 1 2 3) shouldn't match (a #:key x)? Either way, 
I'll file these as a bug.



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