SICP第3.1.1节 - 程序中的本地状态似乎不一致 [英] SICP Section 3.1.1 - Local state in procedures seems inconsistent

问题描述

我正在通过SICP工作。我在第3.1.1节和查看在本地状态。我在GNU Guile v2.0.11中评估这些练习。

I am working my way through SICP. I am on Section 3.1.1 and looking at local state. I am evaluating these exercises in GNU Guile v2.0.11.

我发现了一个关于这一部分的类似问题，但似乎没有解决我所困扰的一点，或者我过于晦涩。

I did find a similar question about this section, but it seems not to address the point I am struggling with, or I am overly obtuse.

我在看的两个例子是：

(define new-withdraw
  (let ((balance 100))
    (lambda (amount)
      (if (>= balance amount)
          (begin (set! balance (- balance amount))
                 balance)
          "Insufficient funds"))))


(define (make-withdraw balance)
  (lambda (amount)
    (if (>= balance amount)
        (begin (set! balance (- balance amount))
               balance)
        "Insufficient funds")))

当我将第一个赋值给一个变量时：

When I assign the first to a variable with:

(define a new-withdraw)
(define b new-withdraw)

我得到两个指向同一个过程对象的指针，它们之间的状态是共享的：

I get two pointers to the same procedure object and the state is shared between them:

scheme@(guile-user)> a 
$1 = #<procedure 1165880 at /path/to/file (amount)>
scheme@(guile-user)> b
$2 = #<procedure 1165880 at /path/to/file (amount)>
scheme@(guile-user)> (a 50)
$3 = 50
scheme@(guile-user)> (b 10)
$4 = 40

当我实现第二个过程时，获取指向具有不同状态的两个不同过程对象的指针：

When I implement the second procedure, though, I get pointers to two distinct procedure objects with distinct state:

scheme@(guile-user)> (define c (make-withdraw 100))
scheme@(guile-user)> (define d (make-withdraw 100))
scheme@(guile-user)> c
$5 = #<procedure 14fdac0 at /path/to/file (amount)>
scheme@(guile-user)> d
$6 = #<procedure 1508360 at /path/to/file (amount)>
scheme@(guile-user)> (c 50)
$7 = 50
scheme@(guile-user)> (d 10)
$8 = 90

没有清楚的解释，我在寻找这个部分时遇到麻烦找到任何东西。我明白一般情况下发生的状况，但我不知道这些程序之间的区别是，允许​​一个单一的通用状态，另一个维持本地状态。

I've read through the section and there is no clear explanation of this, and I'm having trouble finding anything when I search for this section. I understand what is happening in terms of state in general, but I don't get what the difference between these procedures is that allows one to have a single universal state, and the other to maintain local state.

为什么第一个定义new-withdraw无法在多个任务中保持本地状态？看来，lambda应该捕获平衡的分配到100每次我们做不同的分配（define new-withdraw）。

Why is the first definition, 'new-withdraw', unable to maintain local state across multiple assignments? It seems that the lambda should be capturing the assignment of balance to 100 every time we make a different assignment (define new-withdraw).

推荐答案

原因是在第二个例子中，每次都返回一个新的闭包，而在第一个例子中，

The reason is that in the second example you return a fresh closure every time, whereas in the first example there is only one closure, so only one state.

如果您希望在第一种情况下与第二种情况下的行为相同，请更改为：

If you want the same behaviour in the first case as you have in the second case, change to:

(define (new-withdraw) ; this is a procedure now
  (let ((balance 100))
    (lambda (amount)
      (if (>= balance amount)
          (begin (set! balance (- balance amount))
                 balance)
          "Insufficient funds"))))

(define a (new-withdraw)) ; called as a procedure
(define b (new-withdraw))

>

then

> (a 50)
50
> (b 10)
90

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