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Scheme, SICP, R5RS, why is delay not a special form?

Tags:

scheme

sicp

r5rs

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

(cons-stream 1 (display 'hey))

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 >)

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)))

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

like image 231
KnowsLittle Avatar asked Apr 10 '11 08:04

KnowsLittle


1 Answers

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))

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

(define ones (cons-stream 1 ones))

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))]))
like image 99
Eli Barzilay Avatar answered Sep 19 '22 22:09

Eli Barzilay