Aaron W. Hsu wrote:
> This is true in general, but I think the assertion was that two
> procedures that do not need closures to run correctly, can only be
> observationally different or distinct if they have different code bodies.
> I am not seeing a fault in this reasoning off the top of my head.
You are speaking as though whether a procedure needs a closure
to run correctly is a simple binary choice. It is not.
The following example is almost exactly the same as the example
I gave in my post of 9 May:
(define (f m n)
(define (g i)
(define (h j)
(if (= j 0)
i
(g (- j 1))))
(if (< i 0)
(if (= i m)
h ; must a closure be allocated here?
#f)
(h (- i 1))))
(g n))
With the R6RS semantics, a compiler is allowed to rewrite the
escaping reference to h as (lambda (j) (h j)), which means the
closure will be allocated only if i<0 and i=m. With that
rewrite, f allocates no storage at all if its second argument
is non-negative.
With the R6RS semantics, most calls to f are unlikely to need
closures, yet the code vector alone is inadequate to distinguish
proc1 and proc2 defined via
(define proc1 (f -1 -1))
(define proc2 (f -2 -2))
With the R5RS semantics, closures (or unique tags, which are
almost as expensive as closures) will probably have to be
allocated for every call to f and for every internal call to g.
Since most calls to g occur within a tight loop, creating and
garbage-collecting those closures or unique tags is likely to
dominate the running time. That's a high price to pay for a
theoretically inelegant misfeature that's seldom used correctly
in portable code.
Will
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