[racket] macro expansion function in Scheme
Since Ryan's stepper does that and renders the expression for you, I suspect he could supply something close to syntax->datum*.
On Jul 12, 2011, at 10:30 AM, Bas Steunebrink wrote:
> 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.