On Thu 16 Feb 2012 02:29, Noah Lavine <[email protected]> writes:
> (let* ((x (random))
> (y x))
> (eq? x y))
>
> The patch attached to this message lets peval optimize that to
>
> (begin (random) #t)
Neat :)
Note, we don't need to add extra identities to operands: they already
have their gensyms. So to get the effect of this patch, you could add a
clause to fold-constants:
((primcall src 'eq? (lexical _ _ x) (lexical _ _ y))
(if (eq? x y)
(make-const src #t)
<keep-original-expression>))
This works because peval already turns it into the following, after
processing the operands for value:
(let ((x (random)))
(eq? x x))
And all we have to do is check if the lexical is being compared against
itself.
Then, I was about to say:
Now, while this is a valid reduction:
(lambda (x) (eq? x x)) => (lambda (x) #t)
This is not:
(lambda (x) (eqv? x x)) =/> (lambda (x) #t)
The reason is +nan.0: (eqv? +nan.0 +nan.0) => #f
But.... that's wrong! (eqv? +nan.0 +nan.0) isn't specified by the R6RS,
and currently we have it be #t. I guess that's good for our peval
purposes!
Cheers,
Andy
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