lambda - 小阴谋家 : stuck on multiinsertLR&co example

Question

经过一些工作后，我能够理解 multirember&co 函数，但我无法真正理解以下 multiinsertLR&co 代码(第 143 页):

(define multiinsertLR&co
  (lambda (new oldL oldR lat col)
    (cond
     ((null? lat)
      (col '() 0 0))
     ((eq? (car lat) oldL)
      (multiinsertLR&co
       new
       oldL
       oldR
       (cdr lat)
       (lambda (newlat L R) 
         (col (cons new
                    (cons oldL newlat))
              (add1 L) R))))
     ((eq? (car lat) oldR)
      (multiinsertLR&co
       new
       oldL
       oldR
       (cdr lat)
       (lambda (newlat L R)
         (col (cons oldR
                    (cons new newlat))
              L (add1 R)))))
     (else
      (multiinsertLR&co
       new
       oldL
       oldR
       (cdr lat)
       (lambda (newlat L R)
         (col (cons (car lat)
                    newlat) L R)))))))

这本书似乎没有解释评估函数时应该首先通过哪个 collector，所以我使用了 a-friend 收集器和 last- friend 收藏家分别在第 138 页和第 140 页进行了解释。使用任一收集器评估函数会导致以下错误(使用带有 petit chez 方案的跟踪函数):

>> (multiinsertLR&co 'salty 'fish 'chips '(chips and fish or fish and chips) last-friend)

|(multiinsertLR&co salty fish chips (chips and fish or fish and chips)
   #<procedure>)
|(multiinsertLR&co salty fish chips (and fish or fish and chips)
   #<procedure>)
|(multiinsertLR&co salty fish chips (fish or fish and chips) #<procedure>)
|(multiinsertLR&co salty fish chips (or fish and chips) #<procedure>)
|(multiinsertLR&co salty fish chips (fish and chips) #<procedure>)
|(multiinsertLR&co salty fish chips (and chips) #<procedure>)
|(multiinsertLR&co salty fish chips (chips) #<procedure>)
|(multiinsertLR&co salty fish chips () #<procedure>)
Exception: incorrect number of arguments to #<procedure>
Type (debug) to enter the debugger.

代码看了好几遍，都没有发现错误。如果有人有任何见解，请分享。如果有人还可以用一些很好的例子向我指出(相对而言)对延续的温和解释，那也将不胜感激。

Answer 1

这是我解决 Chris 的第二个示例的过程，因为我认为这可能对其他可能被卡住的人有所帮助。 (如有错误请指正!)

;; Anonymous functions (collectors) in mutliinsertLR&co here defined as named  
;; functions for reference purposes
(define left-col
  (lambda (newlat L R)
    (col (cons new
               (cons oldL newlat))
         (add1 L) R)))              ; L counter gets incremented

(define right-col
  (lambda (new lat L R)
    (col (cons oldR
               (cons new newlat))
         L (add1 R))))              ; R counter gets incremented

(define else-col
  (lambda (newlat L R)
    (col (cons (car lat)
               (newlat) L R))))     ; neither counter gets incremented

;; Let's step through an example to see how this works:
;;
;; 1. ENTRY TO FUNCTION
;; new  = foo
;; oldL = bar
;; oldR = baz
;; lat  = (goo bar baz qux quux)
;; col  = list
;;
;; 1. Is lat null? No.
;; 2. Is car of lat equal to oldL? No.
;; 3. Is car of lat equal to oldR? No.
;; 4. Else? Yes. Recur on the function passing the new, oldL
;;    oldR, (cdr lat) and the new collector.
;;
;; 2. FIRST RECURSIVE STEP
;; new  = foo
;; oldL = bar
;; oldR = baz
;; lat  = (bar baz qux quux)
;; col  = else-col
;;
;; - Is lat null? No.
;; - Is car of lat equal to oldL? Yes. Recur on the function
;;   passing new, oldL, oldR, (cdr lat), and the new col.
;;
;; 3. SECOND RECURSIVE STEP
;; new  = foo
;; oldL = bar
;; oldR = baz
;; lat  = (baz qux quux)
;; col  = left-col
;;
;; - Is lat null? No.
;; - Is car of lat equal to oldL? No.
;; - Is car of lat equal to oldR? Yes. Recur on the function
;;   passing new, oldL, oldR, (cdr lat), and the new col.
;;
;; 4. THIRD RECURSIVE STEP
;; new  = foo
;; oldL = bar
;; oldR = baz
;; lat  = (qux quux)
;; col  = right-col
;;
;; - Is lat null? No.
;; - Is car of lat equal to oldL? No.
;; - Is car of lat equal to oldR? No.
;; - Else? Yes. Recur on the function passing new, oldL, oldR,
;;   (cdr lat) and the new collector.
;;
;; 5. FOURTH RECURSIVE STEP
;; new  = foo
;; oldL = bar
;; oldR = baz
;; lat  = (quux)
;; col  = else-col
;;
;; - Is the lat null? No.
;; - Is the car of lat equal to oldL? No.
;; - Is the car of lat equal to oldR? No.
;; - Else? Yes. Recur on the function passing new, oldL, oldR,
;;   (cdr lat) and the new collector.
;;
;; 6. FIFTH RECURSIVE STEP
;; new  = foo
;; oldL = bar
;; oldR = baz
;; lat  = ()
;; col  = else-col
;;
;; - Is the lat null? Yes. Call else-col, where newlat is (),
;;   L is 0 and R is 0.
;;
;; 7. ELSE-COL
;; newlat = ()
;; L      = 0
;; R      = 0
;; col    = else-col
;;
;; - Now call else-col on the (cons (car lat)), which is currently
;;   stored in memory as the atom "quux", onto newlat, as well as
;;   L and R.
;;
;; 8. ELSE-COL (again)
;; newlat = (quux)
;; L      = 0
;; R      = 0
;; col    = right-col
;;
;; - Now call right-col on the (cons (car lat)), which is currently
;;   stored in memory as the atom "qux", onto newlat, as well as L
;;   and R.
;;
;; 9. RIGHT-COL
;; newlat = (qux quux)
;; L      = 0
;; R      = 0
;; col    = left-col
;;
;; - Now call left-col on the cons of oldR and (cons new newlat),
;;   as well as L and the increment of R.
;;
;; 10. LEFT-COL
;; newlat = (baz foo qux quux)
;; L      = 0
;; R      = 1
;; col    = else-col
;;
;; - Now call else-col on the cons of new and (cons oldL newlat),
;;   as well as the increment of L and R.
;;
;; 11. ElSE-COL (last time)
;; newlat = (foo bar baz foo qux quux)
;; L      = 1
;; R      = 1
;; col    = list
;;
;; - Now we call list on the (cons (car lat)), which is currently
;;   stored in memory as the atom "goo", onto newlat, as well as L
;;   and R.
;; - This gives us the final, magical list:
;;   ((goo foo bar baz foo qux quux) 1 1)
;;
;; 12. FIFTH RECURSIVE STEP (redux)
;; new  = foo
;; oldL = bar
;; oldR = baz
;; lat  = (quux)
;; col  = else-col
;;
;; - Base case evaluated, with the result being
;;   ((goo foo bar baz foo qux quux) 1 1).
;; - Function ends.
;;
;; THE END