On Wed, 4 May 2011, Andrzej wrote:
On Wed, May 4, 2011 at 12:07 PM, Andre van Tonder <[email protected]> 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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