Isn’t this because of hygienity in general? For example:
#lang racket
(define-syntax (foo stx)
(syntax-case stx ()
[(_) #'(define doubleup 1)]))
(foo)
doubleup ; unbound id
doesn’t define doubleup.
While the below code does:
#lang racket
(define-syntax (foo stx)
(syntax-case stx ()
[(_)
(with-syntax ([name (datum->syntax stx 'doubleup)])
#'(define name 1))]))
(foo)
doubleup
On Wed, Mar 6, 2019 at 7:57 PM David Storrs <david.sto...@gmail.com> wrote:
> I would like to write a macro that accepts an optional keyword
> argument. If the argument is missing, or if it is present and has a
> true value, then some code will be generated. Otherwise, nothing is
> done. This was my best attempt:
>
> #lang racket
> (require (for-syntax syntax/parse))
> (define-syntax (foo stx)
> (syntax-parse stx
> [(foo name val (~optional (~seq #:make-double? make-double?:boolean)))
> (with-syntax ([func (if (syntax->datum #'(~? make-double? #'#t))
> #'(define (doubleup) (* 2 val))
> #'(begin))])
> #'func)]))
>
> (foo x 6 #:make-double? #t)
> doubleup
>
>
> No joy. doubleup is never defined, no matter what happens with the
> keyword. I'm not sure what I'm doing wrong -- any advice?
>
> More generally, this feels like something for which there should be a
> simpler solution. I think that ~? isn't quite what I'm looking for,
> since the thing that I'm testing for defined/truth isn't going to be
> used in the code that will actually be generated. What should I be
> using?
>
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