(define data (for/list ([x 100000]) x)) (time (begin0 (void) (run2 data))) (time (begin0 (void) (run1 data)))
3 times run1, then 3 times run2, then again 3 times run1. Results are stable. Mon, 16 Jun 2014 08:28:10 +0100 от Matthew Flatt <mfl...@cs.utah.edu>: >I'd expect them to run nearly the same due to inlining and constant >propagation. If I save your program to "ex.rkt" and use > > raco make ex.rkt > raco decompile ex.rkt > >the the output looks almost the same for both functions. > >There's a lot of allocation in these programs, of course, and that's >going to make benchmarking relatively tricky. How are you timing the >functions, and does it matter whether you `run1` or `run2` first? > >At Mon, 16 Jun 2014 11:16:25 +0400, Roman Klochkov wrote: >> Strange. >> >> #lang racket >> (define (test1 x y) >> (if x >> (+ y 1) >> (- y 1))) >> (define (test2 x) >> (if x >> (λ (y) (+ y 1)) >> (λ (y) (- y 1)))) >> (define (run1 data) >> (map (λ (x) (test1 #t x)) data)) >> (define (run2 data) >> (map (λ (x) ((test2 #t) x)) data)) I expect, that run2 should be faster, >> because (test2 #t) returns const (lambda (y) (+ y 1)) and shouldn't be >> checked >> on every iteration. >> >> But in reality (time ...) gives 219 for run1 and 212 for run2. run2 is 1.5 >> times slower! >> >> Why so? >> >> >> -- >> Roman Klochkov____________________ >> Racket Users list: >> http://lists.racket-lang.org/users -- Roman Klochkov
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