Teach Yourself Scheme in Fixnum Days 13 Jump跳转

scheme 一个显著的特性是其支持跳转<Jump>或非局部控制<nonlocal contral>。具体地说，scheme 允许程序控制流跳转到程序的任意地方，不像通过条件语句和函数调用等控制跳转那么局限。

scheme 跳转操作是通过 call-with-current-continuation <简写形式：call/cc>过程来实现的。下面我们将会看到如何通过该过程来实现许多惊人的控制流程<control idioms>。

13.1 call‑with‑current‑continuation

The operator call‑with‑current‑continuation call s its argument, which must be a unary procedure, with a value called the “ current continuation ”. If nothing else, this explains the name of the operator. But it is a long name, and is often abbreviated call/cc . ¹

The current continuation at any point in the execution of a program is an abstraction of the rest of the program . Thus in the program

(+ 1 (call/cc

       (lambda (k)

         (+ 2 (k 3)))))

the rest of the program, from the point of view of the call/cc -application, is the following program-with-a-hole (with [] representing the hole):

call/cc 过程的 "the rest of the program" 就是下面的有一个 [] 的语句：

(+ 1 [])

In other words, this continuation is a program that will add 1 to whatever is used to fill its hole.

This is what the argument of call/cc is called with . Remember that the argument of call/cc is the procedure

(lambda (k)

  (+ 2 (k 3)))

This procedure’s body applies the continuation (bound now to the parameter k ) to the argument 3 . This is when the unusual aspect of the continuation springs to the fore. The continuation call abruptly（突然地） abandons its own computation and replaces it with the rest of the program saved in k ! In other words, the part of the procedure involving the addition of 2 is jettisoned（抛弃）, and k ’s argument 3 is sent directly to the program-with-the-hole:

(+ 1 [])

The program now running is simply

(+ 1 3)

which returns 4 . In sum,

(+ 1 (call/cc

       (lambda (k)

         (+ 2 (k 3)))))

=>  4

The above illustrates what is called an escaping continuation, one used to exit out of a computation (here: the (+ 2 []) computation). This is a useful property, but Scheme’s continuations can also be used to return to previously abandoned contexts, and indeed to invoke them many times. The “rest of the program” enshrined in a continuation is available whenever and how many ever times we choose to recall it, and this is what contributes to the great and sometimes confusing versatility of call/cc . As a quick example, type the following at the listener:

(define r #f)

(+ 1 (call/cc

       (lambda (k)

         (set! r k)

         (+ 2 (k 3)))))

=>  4

The latter expression returns 4 as before. The difference between this use of call/cc and the previous example is that here we also store the continuation k in a global variable r .

Now we have a permanent record of the continuation in r . If we call it on a number, it will return that number incremented by 1 :

(r 5)

=>  6

Note that r will abandon its own continuation, which is better illustrated by embedding the call to r inside some context:

而把 r 调用嵌入在某些环境中更能证明 continuation 跳转的作用：

(+ 3 (r 5))

=>  6

r 调用会将程序控制流直接跳转到 "the continuation" (+ 1 []) 处，再用 5 替换 []，执行那里的代码。

The continuations provided by call/cc are thus abortive continuations.

13.2  Escaping continuations

Escaping continuations are the simplest use of call/cc and are very useful for programming procedure or loop exits. Consider a procedure list‑product that takes a list of numbers and multiplies them. A straightforward recursive definition for list‑product is:

(define list-product

  (lambda (s)

    (let recur ((s s))为什么，不要不行

      (if (null? s) 1

          (* (car s) (recur (cdr s)))))))

There is a problem with this solution. If one of the elements in the list is 0 , and if there are many elements after 0 in the list, then the answer is a foregone 预知conclusion. Yet, the code will have us go through many fruitless recursive calls to recur before producing the answer. This is where an escape continuation comes in handy. Using call/cc , we can rewrite the procedure as:

但是上面的计算方法有一个问题，就是列表中有一个元素为 0 并且其后还有许多个元素，当计算到 0 时，结果就可以确定为 0，而不用在与其后的元素相乘。此时，我们可以用 escaping continuations 重写 list-product 过程来避免这些无效的计算：

(define list-product

  (lambda (s)

    (call/cc

      (lambda (exit)

        (let recur ((s s))

          (if (null? s) 1

              (if (= (car s) 0) (exit 0)

                  (* (car s) (recur (cdr s))))))))))

If a 0 element is encountered, the continuation exit is called with 0 , thereby avoiding further calls to recur .

13.3  Tree matching

A more involved example of continuation usage is the problem of determining if two trees (arbitrarily nested dotted pairs) have the same fringe , ie, the same elements (or leaves ) in the same sequence. Eg,

先序排列是否相同

(same-fringe? '(1 (2 3)) '((1 2) 3))

=>  #t

(same-fringe? '(1 2 3) '(1 (3 2)))

=>  #f

The purely functional approach is to flatten both trees and check if the results match.

一个简单的方法就是先计算出两个树的先序排列表再比较它们

define same-fringe?

  (lambda (tree1 tree2)

    (let loop ((ftree1 (flatten tree1))

               (ftree2 (flatten tree2)))

      (cond ((and (null? ftree1) (null? ftree2)) #t)

            ((or (null? ftree1) (null? ftree2)) #f)

            ((eqv? (car ftree1) (car ftree2))

             (loop (cdr ftree1) (cdr ftree2)))

            (else #f)))))

(define flatten

  (lambda (tree)

    (cond ((null? tree) '())

          ((pair? (car tree))

           (append (flatten (car tree))

                   (flatten (cdr tree))))

          (else

           (cons (car tree)

                 (flatten (cdr tree)))))))

However, this traverses the trees completely to flatten them, and then again till it finds non-matching elements. Furthermore, even the best flattening algorithms will require cons es equal to the total number of leaves. (Destructively modifying the input trees is not an option.)

We can use call/cc to solve the problem without needless traversal and without any cons ing. Each tree is mapped to a generator , a procedure with internal state that successively produces the leaves of the tree in the left-to-right order that they occur in the tree.

(define tree->generator

  (lambda (tree)

    (let ((caller '*))

      (letrec

          ((generate-leaves

            (lambda ()

              (let loop ((tree tree))

                (cond ((null? tree) 'skip)

                      ((pair? tree)

                       (loop (car tree))

                       (loop (cdr tree)))

                      (else

                       (call/cc

                        (lambda (rest-of-tree)

                          (set! generate-leaves

                            (lambda ()

                              (rest-of-tree 'resume)))

                          (caller tree))))))

              (caller '()))))

        (lambda ()

          (call/cc

           (lambda (k)

             (set! caller k)

             (generate-leaves))))))))

When a generator created by tree‑>generator is called, it will store the continuation of its call in caller , so that it can know who to send the leaf to when it finds it. It then calls an internal procedure called generate‑leaves which runs a loop traversing the tree from left to right. When the loop encounters a leaf, it will use caller to return the leaf as the generator’s result, but it will remember to store the rest of the loop (captured as a call/cc continuation) in the generate‑leaves variable. The next time the generator is called, the loop is resumed where it left off so it can hunt for the next leaf.

Note that the last thing generate‑leaves does, after the loop is done, is to return the empty list to the caller . Since the empty list is not a valid leaf value, we can use it to tell that the generator has no more leaves to generate.

The procedure same‑fringe? maps each of its tree arguments to a generator, and then calls these two generators alternately. It announces failure as soon as two non-matching leaves are found:

(define same-fringe?

  (lambda (tree1 tree2)

    (let ((gen1 (tree->generator tree1))

          (gen2 (tree->generator tree2)))

      (let loop ()

        (let ((leaf1 (gen1))

              (leaf2 (gen2)))

          (if (eqv? leaf1 leaf2)

              (if (null? leaf1) #t (loop))

              #f))))))

It is easy to see that the trees are traversed at most once, and in case of mismatch, the traversals extend only upto the leftmost mismatch. cons is not used.

一个更加容易理解的版本：

http://lispor.is-programmer.com/posts/23105.html

更多：

http://www.ccs.neu.edu/home/dorai/t-y-scheme/t-y-scheme-Z-H-15.html#node_chap_13

http://lispor.is-programmer.com/posts/23638.html