Do Racket streams memoize their elements?

问题

Does Racket use memoization when computing large amounts of numbers from an infinite stream? So, for example, if I printed out (aka, computed and displayed) the first 400 numbers on the infinite stream of integers: (1 2 3 ... 399 400) And right after I asked to print the first 500 numbers on this infinite stream. Would this second set of computations use memoization? So the first 400 numbers would not be computed again?

Or does this functionality need to be coded by the user/obtained from libraries?

回答1:

The built-in racket/stream library uses lazy evaluation and memoization to draw elements from a stream:

(require racket/stream)

(define (print-and-return x)
  (displayln "drawing element...")
  x)

(define (in-range-stream n m)
  (if (= n m)
      empty-stream
      (stream-cons (print-and-return n) (in-range-stream (add1 n) m))))

(define s (in-range-stream 5 10))

(stream-first s)
(stream-first s)
(stream-first (stream-rest s))

The expressions passed to stream-cons are not evaluated until requested either with stream-first or stream-rest. Once evaluated, they are memoized. Notice that despite the four stream operations performed on s, only two `"drawing element..." messages are displayed.

回答2:

You can use the memoize package.

GitHub source: https://github.com/jbclements/memoize/tree/master

raco pkg install memoize

Using it is as simple as replacing define with define/memo. To quote its example:

(define (fib n)                                     
  (if (<= n 1) 1 (+ (fib (- n 1)) (fib (- n 2)))))  

> (time (fib 35))                                   
cpu time: 513 real time: 522 gc time: 0             
14930352                                            

> (define/memo (fib n)                              
    (if (<= n 1) 1 (+ (fib (- n 1)) (fib (- n 2)))))

> (time (fib 35))                                   
cpu time: 0 real time: 0 gc time: 0                 
14930352

Also, it is generally quite easy to implement memoization yourself using a Racket hash-table.

回答3:

For other Scheme implementations that use SRFI 41 streams, those streams also fully memoise all the materialised elements.

In fact, in my Guile port of SRFI 41 (which has been in Guile since 2.0.9), the default printer for streams will print out all the elements so materialised (and nothing that isn't):

scheme@(guile-user)> ,use (srfi srfi-41)
scheme@(guile-user)> (define str (stream-from 0))
scheme@(guile-user)> (stream-ref str 4)
$1 = 4
scheme@(guile-user)> str
$2 = #<stream ? ? ? ? 4 ...>

Any of the elements that aren't being printed out as ? or ... have already been memoised and won't be recomputed. (If you're curious about how to implement such a printer, here's the Guile version.)

来源：https://stackoverflow.com/questions/26478167/do-racket-streams-memoize-their-elements

标签

stream

racket