My message below is not doing the right test. This is the right test
and it produces the expected result, namely that the TR version runs a
bit faster (presumably because of type-based optimizations that TR
does).
Robby
#lang racket
(module t typed/racket/base
(: divides? : Integer Integer -> Boolean)
(define (divides? a b)
(cond [(zero? a) #f]
[else (= (remainder b a) 0)]))
(provide divides?))
(module c racket/base
(require racket/contract/base)
(define (divides? a b)
(cond [(zero? a) #f]
[else (= (remainder b a) 0)]))
(provide
(contract-out
[divides? (-> exact-integer? exact-integer? boolean?)])))
(require (prefix-in c: (submod "." c)))
(require (prefix-in t: (submod "." t)))
(time
(for ([x (in-range 3 5000000)])
(if (3 . c:divides? . x)
x
0)))
(time
(for ([x (in-range 3 5000000)])
(if (3 . t:divides? . x)
x
0)))
On Mon, Aug 24, 2015 at 8:18 AM, Robby Findler
<[email protected]> wrote:
> It looks to me like the slowdown isn't entirely explained by contract
> checking, or perhaps TR isn't generating the contracts I would have
> guessed. With the program below, I see this output
>
> cpu time: 1228 real time: 1228 gc time: 133
> cpu time: 658 real time: 658 gc time: 18
> cpu time: 80 real time: 81 gc time: 0
>
> but would have expected the first two lines to be nearly the same.
>
> Robby
>
> #lang racket
>
> (require (only-in math/number-theory divides?))
>
> (define (divisible-by? x d)
> (= (modulo x d)
> 0))
>
> (module d racket/base
> (require racket/contract/base)
> (define (divisible-by? x d)
> (= (modulo x d)
> 0))
> (provide
> (contract-out
> [divisible-by? (-> exact-integer? exact-integer? boolean?)])))
>
> (require (prefix-in c: (submod "." d)))
>
>
> (module+ main
> (time
> (for ([x (in-range 3 5000000)])
> (if (3 . divides? . x)
> x
> 0)))
>
> (time
> (for ([x (in-range 3 5000000)])
> (if (3 . c:divisible-by? . x)
> x
> 0)))
>
> (time
> (for ([x (in-range 3 5000000)])
> (if (x . divisible-by? . 3)
> x
> 0))))
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