方案，SICP，R5RS，为什么延迟不是一种特殊形式? [英] Scheme, SICP, R5RS, why is delay not a special form?

问题描述

这与SICP的第3.5章有关，其中正在讨论流.这个想法是:

This is concerning chapter 3.5 from SICP, in which streams are being discussed. The idea is that:

(cons-stream 1 (display 'hey))

不应评估cons流的第二部分，因此不应打印嘿".确实发生了，我得到以下输出:

Should not evaluate the second part of the cons-stream, so it should not print "hey". This does happen, I get the following output:

hey(1.#< promise>)

所以我的结论是，延迟不是以特殊形式实现的吗?还是我做错了什么?我使用以下实现:

So my conclusion is that delay is not implemented as a special form? Or am I doing something wrong? I use the following implementation:

(define (cons-stream a b) 
  (cons a (delay b)))

延迟是默认的R5RS实现.这是实现上的错误吗，还是我没有做或不正确理解它?

With delay being the default R5RS implementation. Is this a fault in the implementation, or am I not doing or understanding it right?

推荐答案

您做创建了一个promise，但是该promise是在您的cons-stream内部创建的，这意味着为时已晚，并且该表达式已被评估.试试这个:

You do create a promise, but the promise is created inside your cons-stream, which means that it's too late and the expression was already evaluated. Try this:

(define (foo x)
  (display "foo: ") (write x) (newline)
  x)

(cons-stream 1 (foo 2))

，您将发现评估还为时过早.出于同样的原因，这是

and you'll see that it's evaluated too early. For the same reason, this:

(define ones (cons-stream 1 ones))

当cons-stream是一个函数时，

和任何其他无限列表将不起作用.因此，事实是delay是一种特殊形式，但是您没有使用它的功能，因为您将cons-stream定义为普通函数.如果要使 it 以相同的特殊方式运行，则必须将cons-stream定义为宏.例如:

and any other infinite list won't work when your cons-stream is a function. So the thing is that delay is a special form, but you're not using its feature since you define cons-stream as a plain function. You have to define cons-stream as a macro if you want to make it behave in the same special way too. For example:

(define-syntax cons-stream
  (syntax-rules ()
    [(cons-stream x y) (cons x (delay y))]))

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