Thanks Matthias and Eli, I understand what expand does and that
syntax->datum throws away too much (by design). Still my question
remains of how to obtain a datum that is equivalent to a given quoted
expression (when eval'ed) but with all macros expanded (generating fresh
symbols where necessary). So does there exist a function syntax->datum*
such that:
> (define-syntax m
(syntax-rules ()
((_ x)
(lambda (x) (lambda (y) x)))))
> (((m y) #t) #f)
#t ; OK
> (syntax->datum* (expand '(m y)))
(lambda (y) (lambda (g42) y)) ; g42 is a fresh symbol
> (((eval (syntax->datum* (expand '(m y)))) #t) #f)
#t ; using Racket's syntax->datum gives #f
Please note that I'm really interested in getting just the
macro-expanded datum (otherwise I would've been done already!).
Thanks again for your kind help,
Bas
On 7/12/11 16:17 PM, Matthias Felleisen wrote:
On Jul 12, 2011, at 10:16 AM, Eli Barzilay wrote:
8 minutes ago, Matthias Felleisen wrote:
The following experiment produces a bit more insight:
(((eval (expand '(m y))) #t) #f)
#t
The result of expand compiles to the correct closure.
So the docs for syntax->datum are dead serious when they say that it
"returns a datum by stripping the lexical information, source
location information, properties, and tamper status from stx."
Try also the macro stepper. From an upcoming package for the repl:
-> (define-syntax m
(syntax-rules ()
((_ x)
(lambda (x) (lambda (y) x)))))
-> ,stx (m y) *
syntax set
stepper:
Macro transformation
(m y)
==>
(lambda:1 (y) (lambda:1 (y:1) y))
(But you'd usually want to use the gui version...)
Fire up drracket and run the stepper.
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