Re: [Scheme-reports] ANN: first draft of R7RS small language available
Andre van Tonder 04 May 2011 13:30 UTC
On Wed, 4 May 2011, Andrzej wrote:
> On Wed, May 4, 2011 at 12:07 PM, Andre van Tonder <andre@het.brown.edu> wrote:
>>
>> Be that as it may, let me give another example. I am pretty sure that the
>> the spec requires the following to evaluate to 1.
>>
>> (define-syntax nonfalse-identity
>> (syntax-rules ()
>> ((_ x)
>> (cond (x => (lambda (x) x))))))
>>
>> (let ((else 1))
>> (nonfalse-identity else)) ====> 1
>>
>> but I think your implementation will give the wrong result here (or an
>> error).
>
> You're right. An implementation conforming to R5RS should fail here.
> '=>' is not allowed in the 'else' clause.
No, I claimed that a correct implementation must sucxceed and give 1. Macro
hygiene reequires it to. See the explanation below.
>> Here is anotehr example, which evaluates a given expression (in case it has
>> side effects) and returns 1. It is a silly way of doing this, but there is
>> no doubt that the spec requires the answer to be 1.
>>
>> (define-syntax map-to-identity
>> (syntax-rules ()
>> ((_ exp)
>> (cond (#t exp 1)))))
>>
>> (let ((=> 0))
>> (map-to-identity =>)) ======> 1
>
> Ditto. A clause with '=>' must contain a single operand procedure (not '1').
Again, I calim that teh spec requires the answer to be 1, again because of
macro hygiene. I would suggest that you reread the section on macro expansion:
If a macro transformer inserts a binding for an identifier (variable or
keyword), the identifier will in effect be renamed throughout its scope to
avoid conflicts with other identifiers.
In other words, the => in
(let ((=> 0))
(map-to-identity =>))
is renamed during expansion to become (a lambda expression) equivalent to
(let ((g0 0))
(map-to-identity g0)
where g0 is a generated identifier, all before MAP-TO-IDENTITY is even
encountered. This then expands to the equivalent of
(let ((g0 0))
(cond (#t g0 1)))
which evaluates to 1.
So the relevant part of the spec that you were missinbg had to do with the
expansion algorithm.
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