Question

SICP书的摘录，对于给出的最后一行，方案代码编译存在问题，在我看来，函数sqr和shower都是返回单个值的单个参数的函数。因此应该工作。对于以下给定的方案代码：

(define (stream-car stream)
  (car stream))

(define (stream-cdr stream)
  (force (cdr stream)))

;; cons-stream needs to be a macro

(define-syntax cons-stream
  (syntax-rules ()
    [(_ a b) (cons a (delay b))]
    ))

(define the-empty-stream '())

(define stream-null? null?)

;; define stream-for-each
;; run a procedure on each element of stream
(define (stream-for-each proc stream)
  (if (stream-null? stream)
      'done
      (begin (proc (stream-car stream))
             (stream-for-each proc (stream-cdr stream)))))

(define test-stream (cons-stream 2 (cons-stream 3 (cons-stream 4 '()))))

;; displays passed argument but with a newline
(define (display-line x)
  (newline)
  (display x))

(define (display-stream stream)
  (stream-for-each display-line stream))

;; stream map
(define (stream-map proc stream)
  (if (stream-null? stream)
      the-empty-stream
      (cons-stream (proc (stream-car stream))
                   (stream-map proc (stream-cdr stream)))))

(define (sqr x)
  (* x x))

(define (stream-enumerate-interval low hi)
  (if (> low hi)
      '()
      (cons-stream low (stream-enumerate-interval (+ 1 low) hi))))

(define (shower x)
  (display-line x)
  x)

;; this works
(define yyy (stream-map sqr (stream-enumerate-interval 1 10)))


;; reason why below wont work?
;; (define xxx (stream-map shower (stream-enumerate-interval 1 10)))

同样根据SICP一书，它定义了一个不同版本的流映射，它允许接受可以接受多个参数的过程的过程。

我无法理解是否需要不同的流映射以及为什么上一行的流映射代码无法正常工作

Answer 1

它们都有效，但由于stream-map是懒惰的，除非你做一些强迫它，否则你不会完全计算它们。例如。让我们stream->list：

(define (stream->list stream)
  (if (stream-null? stream)
      '()
      (cons (stream-car stream)
            (stream->list (stream-cdr stream)))))

(stream->list yyy)
; ==> (1 4 9 16 25 36 49 64 81 100)

(stream->list zzz)
; ==> (1 2 3 4 5 6 7 8 9 10) (and as a side effect prints then one line at a time)

事实上，除非您尝试访问该值，否则永远不会对列表的其余部分进行实际计算。 stream->list强制计算整个流，然后计算出来。

请注意，sicp流总是强制第一个值。这是其他流库已修复的简化，因此在SRFI-41中除非您访问结果，否则您没有任何副作用。

#!r6rs
(import (rnrs)
        (srfi :41))

;; nothing is displayed after this expression is evaluated 
(define squares (stream-map (lambda (x) (shower (* x x)))
                            (stream-from 0)))

(stream-car squares)
; ==> 0 (displays "\n0")

(stream->list 4 squares)
; ==> (0 1 4 9) 
; (displayes "\n1\n4\n9" since "0" is already calculated.)

方案简单的流映射在SICP示例中不起作用

1 个答案: