Could you put the dialect values in separate modules and have the #%module-begin conditionally pick which dialect modules to require?
On Thu, Feb 11, 2016 at 11:18 AM, Matthew Butterick <m...@mbtype.com> wrote: > > Why don't you put the val-id into the module via foo? Why is bar > supposed to do that? > > > This is part of a larger pattern where I've got a basic `#%module-begin` > for my #lang that I want to spin out into dialect-specific versions. The > dialects only vary in certain values. > > Sure, I can make a wrapper macro that generates the `#%module-begin` > macro. That's what I've done in the past. It works. But a) it's unhygienic, > and I'm trying to learn cleaner habits, and b) I was recently reminded [1] > that repeated code should be factored out of macros. > > But I'm starting to think that the specialness of `#%module-begin` resists > this kind of hacking. > > [1] > http://docs.racket-lang.org/style/Language_and_Performance.html#%28part._.Macros__.Space_and_.Performance%29 > > > > On Feb 11, 2016, at 7:08 AM, Matthias Felleisen <matth...@ccs.neu.edu> > wrote: > > > > > Why don't you put the val-id into the module via foo? Why is bar > supposed to do that? > > > > > > > > > > On Feb 10, 2016, at 7:13 PM, Matthew Butterick <m...@mbtype.com> wrote: > > > >>>> You really want to parametrize foo over x for such things. (Or use a > syntax parameter.) > >> > >> > >> Of course, though I see now that my example is perhaps too schematic. > If I were really adding, I could do this: > >> > >> ;;;;;;;;;;;;;; > >> #lang racket > >> (require racket/stxparam rackunit) > >> > >> (define-syntax-parameter x-param > >> (λ (stx) (raise-syntax-error (syntax-e stx) "must be parameterized > first"))) > >> > >> (define-syntax-rule (foo) (+ x-param x-param)) > >> > >> (define-syntax bar > >> (syntax-rules () > >> [(_ val-in) > >> (let ([val-id val-in]) > >> (syntax-parameterize ([x-param (make-rename-transformer #'val-id)]) > >> (foo)))])) > >> > >> (check-equal? (bar 42) 84) > >> ;;;;;;;;;;;;;; > >> > >> Wonderful. But in fact, both my `foo` and `bar` are #%module-begin > transformers. So I can try to follow the same pattern, but I get stuck at > the end: > >> > >> ;;;;;;;;;;;;;; > >> #lang racket > >> (require racket/stxparam) > >> > >> (define-syntax-parameter x-param > >> (λ (stx) (raise-syntax-error (syntax-e stx) "must be parameterized > first"))) > >> > >> (define-syntax-rule (foo expr ...) (#%module-begin x-param expr ...)) > >> > >> (define-syntax bar > >> (syntax-rules () > >> [(_ val-in) > >> (let ([val-id val-in]) > >> (syntax-parameterize ([x-param (make-rename-transformer #'val-id)]) > >> <what goes here?>))])) > >> ;;;;;;;;;;;;;; > >> > >> If I put `(foo)` on the last line, I'm trying to `syntax-parameterize` > over a module-begin context, which fails. > >> > >> AFAICT it's also not possible to syntax-parameterize over > `make-rename-transformer`, because it produces a transformer inside a > transformer, and I'm not clear how to unwrap this so that `bar` is a usable > #%module-begin macro: > >> > >> ;;;;;;;;;;;;;; > >> #lang racket > >> (require racket/stxparam) > >> > >> (define-syntax-parameter x-param > >> (λ (stx) (raise-syntax-error (syntax-e stx) "must be parameterized > first"))) > >> > >> (define-syntax-rule (foo expr ...) (#%module-begin x-param expr ...)) > >> > >> (define-syntax bar > >> (<what goes here?> > >> (λ(stx) > >> #'(let ([val-id 42]) > >> (syntax-parameterize ([x-param (make-rename-transformer > #'val-id)]) > >> (make-rename-transformer #'foo)))))) > >> ;;;;;;;;;;;;;; > >> > >> > >> > >> On Feb 10, 2016, at 2:52 PM, Matthias Felleisen <matth...@ccs.neu.edu> > wrote: > >> > >>> > >>> You really want to parametrize foo over x for such things. (Or use a > syntax parameter.) > >>> > >>> > >>> On Feb 10, 2016, at 3:26 PM, Matthew Butterick <m...@mbtype.com> wrote: > >>> > >>>> I suspect there's a simple way to do this, I just haven't done it > before, so it does not appear simple. I tried a syntax parameter, and then > I had two problems. > >>>> > >>>> Please consider this schematic example (no, I am not using macros to > add): > >>>> > >>>> #lang racket > >>>> (define-syntax-rule (foo) (+ x x)) > >>>> (define-syntax bar (make-rename-transformer #'foo)) > >>>> > >>>> `(foo)` will fail with an undefined-identifier error because there is > no binding for x. As will `(bar)`, because it just refers to `(foo)`. > >>>> > >>>> Q: At the definition site of `bar`, is there a way to bind x so that > it affects `foo`, and `(bar)` thereby produces a meaningful answer? > >>>> > > -- > You received this message because you are subscribed to the Google Groups > "Racket Users" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to racket-users+unsubscr...@googlegroups.com. > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Groups "Racket Users" group. 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