实现产量和计划发送

Question

我试图端口yield和yield from从Python的方案。

这是我做过的实现：

(define (coroutine routine)
  (let ((current routine)
    (status 'new))
    (lambda* (#:optional value)
      (let ((continuation-and-value
         (call/cc (lambda (return)
            (let ((returner
                   (lambda (value)
                 (call/cc (lambda (next)
                        (return (cons next value)))))))
              (if (equal? status 'new)
                  (begin
                (set! status 'running)
                (current returner))
                  (current (cons value returner)))
              (set! status 'dead))))))
    (if (pair? continuation-and-value)
        (begin (set! current (car continuation-and-value))
           (cdr continuation-and-value))
        continuation-and-value)))))

这个问题，这个实现的是，它必须调用方式并不看起来像Python的yield。

(define why (call/cc (lambda (yield)
               (format #t "love me or leave me!")
               (yield "I leave!")
               ;; the program never reach this part
               (format #t "it probably left :("))))
(format #t "return actually populates WHY variable\n")
(format #t "WHY: ~a\n")

别的不说，每次我需要重新启动协程，我必须let新return变量能够exit协程。基本上，我觉得语法太冗长。是否有其他有更清晰的语法？

它应该可以yield和send值的协程。下面是如何的协同程序必须使用一个例子：

(define-coroutine (zrange start step)
  "compute a range of values starting a START with STEP between
   each value. The coroutine must be restarted with 0 or more, which
   is added to the step"
  (let loop ((n start))
    (loop (+ n step (yield n)))))


(coroutine-map (zrange 0 10) '(1 100 1000 10000 100000))
;; => 0 110 1120 11130 111140

在上文中，1被忽略，然后100，1000是send到发电机上。我做了一个实现的基础上，@sylwester代码，但是我有一个宏观的烦恼：

(define (make-generator procedure)
  (define last-return #f)
  (define last-value #f)
  (define last-continuation (lambda (_) (procedure yield)))

  (define (return value)
    (newline)(display "fuuu")(newline)
    (call/cc (lambda (continuation)
               (set! last-continuation continuation)
               (set! last-value value)
               (last-return value))))
  (lambda* (. rest)  ; ignore arguments
    (call/cc (lambda (yield)
               (set! last-return yield)
               (apply last-continuation rest)))))

(define-syntax define-coroutine
  (syntax-rules ()
    ((_ (name args ...) body ...)
     (define (name args ...)

       (make-generator
        (lambda (yield)
          body ...))))))

(define-coroutine (zrange start step)
  (let loop ((n start))
     (loop (+ n step (yield n)))))

(display (map (zrange 0 10) '(1 100 1000 10000 100000)))

Answer 1

事情是这样的：

(define (make-generator procedure)
  (define last-return values)
  (define last-value #f)
  (define (last-continuation _) 
    (let ((result (procedure yield))) 
      (last-return result)))

  (define (yield value)
    (call/cc (lambda (continuation)
               (set! last-continuation continuation)
               (set! last-value value)
               (last-return value))))

  (lambda args
    (call/cc (lambda (return)
               (set! last-return return)
               (if (null? args)
                   (last-continuation last-value)
                   (apply last-continuation args))))))

使用这样的：

(define test 
 (make-generator
   (lambda (collect)
     (collect 1)
     (collect 5)
     (collect 10)
     #f)))

(test) ; ==> 1
(test) ; ==> 5
(test) ; ==> 10
(test) ; ==> #f (procedure finished)

现在，我们可以包装内部成宏：

(define-syntax (define-coroutine stx)
  (syntax-case stx ()
    ((_ (name . args) . body )
     #`(define (name . args)
         (make-generator 
          (lambda (#,(datum->syntax stx 'yield))
            . body))))))

请注意，define-coroutine使用语法的情况，因为我们需要yield不卫生的实现。

(define-coroutine (countdown-from n)
  (let loop ((n n))
    (if (= n 0)
        0
        (loop (- (yield n) 1)))))

(define countdown-from-10 (countdown-from 10))

(define (ignore procedure)
  (lambda ignore
    (procedure)))

(map (ignore countdown-from-10) '(1 1 1 1 1 1)) ; ==> (10 9 8 7 6 5)

;; reset
(countdown-from-10 10)  ; ==> 9
(countdown-from-10)     ; ==> 8
;; reset again
(countdown-from-10 100) ; ==> 99

Answer 2

荣誉对@Sylwester为一个伟大的答案。

困难的部分是让yield提供给发电机的功能。 datum->syntax创建一个语法对象，并要求您提供从为新对象的背景下采取的另一种语法对象。在这种情况下，我们可以使用STX具有相同的上下文传递到宏观的功能。

如果人们发现它有用，我用一个简单的版本：

(define-syntax (set-continuation! stx)
  "Simplifies the common continuation idiom
    (call/cc (λ (k) (set! name k) <do stuff>))"
  (syntax-case stx ()
    [(_ name . body)
     #`(call/cc (λ (k)
                  (set! name k)
                  . body))]))

(define-syntax (make-generator stx)
  "Creates a Python-like generator. 
   Functions passed in can use the `yield` keyword to return values 
   while temporarily suspending operation and returning to where they left off
   the next time they are called."
  (syntax-case stx ()
    [(_ fn)
     #`(let ((resume #f)
             (break #f))
         (define #,(datum->syntax stx 'yield)
           (λ (v)
             (set-continuation! resume
               (break v))))
         (λ ()
           (if resume
               (resume #f)
               (set-continuation! break
                 (fn)
                 'done))))]))

它的用法的例子：

(define countdown
  (make-generator
   (λ ()
     (for ([n (range 5 0 -1)])
           (yield n)))))

(countdown)
=> 5
...
(countdown)
=> 1
(countdown)
=> 'done
(countdown)
=> 'done

Answer 3

一种方法是在这里。如果您使用的诡计，你应该使用提示（他们是约两个数量级比使用与狡诈完全延续更快）：

How to implement Python-style generator in Scheme (Racket or ChezScheme)?