Ray Dillinger (single >) quoting me (>>) quoting John Cowan (>>>):
>>> I'm probably missing something here, but if h doesn't need a
>>> closure, what's wrong with using the code address as a unique
>>> tag?
>> Because that would violate both R5RS and R6RS semantics.
>> Both the R5RS and R6RS insist that procedures that are eqv? behave
>> the same. Distinct procedures that share the same code are
>> unlikely to behave the same.
> Um? Will, did you misspeak here or is there something very
> fundamental that I'm not understanding? As I see it the same
> vector of code ought to produce the same behavior when run.
I did not misspeak. Jeff Read explained my meaning.
Here's a concrete example:
(define (make-counter)
(let ((n 0))
(lambda () (set! n (+ n 1)) n)))
(define f (make-counter))
(define g (make-counter))
The values of f and g are procedures. Those procedures are likely
to be represented by closures that share the same code vector.
With John Cowan's suggested implementation of eqv?, (eqv? f g)
would return true, which would violate both R5RS and R6RS semantics.
> Different nonempty closures over the same procedure are unique,
> of course, but the procedure itself?
A closure is a representation of a procedure. In the various
Scheme mailing lists, most uses of the word "closure" should
really be uses of the word "procedure" instead.
Will
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