实现 yield 并在 Scheme 中发送

Question

我试图端口yield，并yield from从Python的方案。

这是我完成的一个实现：

(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)))))

这个实现的问题在于它必须被调用的方式看起来不像 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")

除此之外，每次我需要重新启动协程时，我必须 let一个新return变量才能exit启动协程。基本上，我觉得语法太冗长了。还有其他语法更简洁的吗？

它应该是可能的yield 和 send值的协程。以下是必须如何使用协程的示例：

(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

在上面,1被忽略然后100,1000是send生成器。我已经基于@sylwester 代码完成了一个实现，但是我在使用宏时遇到了麻烦：

(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

像这样的东西：

(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))))))

像这样使用：

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

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

现在我们可以将内部组件包装成一个宏：

(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，这define-coroutine是使用语法案例实现的，因为我们需要使yield不卫生。

(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