Asumu,
Your lookup macro's output just tells whether there is a rename somewhere
between the binding for f and its original source binding. Rename
transformers get injected all over the place. To get the real story, turn
your lookup macro into a loop that chases the binding back to the source.
Or, use the macro stepper with hiding off. The let-syntax expression
expands into:
(letrec-syntaxes+values (((f1) (make-set!-transformer values))) ()
(letrec-syntaxes+values (((f) (values (make-rename-transformer
(quote-syntax f1))))) ()
(lookup f)))
That's where the rename comes from.
Carl Eastlund
On Fri, May 17, 2013 at 7:30 PM, Asumu Takikawa <as...@ccs.neu.edu> wrote:
> Hi all,
>
> I'm confused about an aspect of set! and rename transformers. I'll
> explain with this example:
>
> #lang racket
>
> ;; a macro that uses `syntax-local-value/immediate`
> (define-syntax (lookup stx)
> (syntax-case stx ()
> [(_ id)
> (let-values ([(f _) (syntax-local-value/immediate #'id)])
> (displayln (rename-transformer? f))
> (displayln (set!-transformer? f))
> #'0)]))
>
> ;; f is a set!-transformer
> (let-syntax ([f (make-set!-transformer values)])
> (lookup f))
>
> ;; sanity check
> (rename-transformer? (make-set!-transformer values))
>
> In this example, `f` is bound to a set!-transformer. The macro `lookup`
> will look up the value bound to `f` at compile-time, and I expected that
> the result would be the set! transformer.
>
> However, it seems like the set! transformer is somehow being turned into
> a rename transformer (note the two print statements produce #t and #f
> respectively). The last line suggests that set! transformers are not
> actually a "subtype" of rename transformers though.
>
> Am I confused about set! transformers or is there a bug here?
>
> Cheers,
> Asumu
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