[racket] macro expansion function in Scheme

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.