Also: Racket's contract system can help you find inputs that
distinguish these functions. Try this out:
#lang racket
(define (f1 x y)
(if (and (zero? x)
(= (+ x y) y))
1
0))
(define (f2 x y)
(if (zero? x)
1
0))
(define/contract (same? x y)
(-> any/c any/c #t)
(equal? (with-handlers ([exn:fail? exn-message])
(f1 x y))
(with-handlers ([exn:fail? exn-message])
(f2 x y))))
(for ([x (in-range 100)])
(contract-exercise same?))
On Tue, Mar 6, 2018 at 8:16 AM, Matthew Flatt <[email protected]> wrote:
> At Fri, 2 Mar 2018 12:34:00 +0100, "'Paulo Matos' via Racket Developers"
> wrote:
>> (define (f x y)
>> (if (and (zero? x)
>> (= (+ x y) y))
>> 1
>> 0))
>>
>> In reality this is the same as:
>> (define (f x y)
>> (if (zero? x)
>> 1
>> 0))
>
> Those turn out to be different. Try `(f 0 +nan.0)` or `(f 0 'symbol)`.
>
>> Except racket does not perform the optimization. What's the reason for
>> this?
>
> Even if you throw in enough constraints on the program to make a
> similar transformation valid, the optimizer would have to include
> specific mechanisms for tracking zero values and number types (more
> than it does) and rules on how arithmetic operations interact with
> those abstract values. We haven't so far had a reason to create all of
> those rules. And as your example illustrates, they'd help less often
> than you might expect.
>
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