what is the difference between these two Scheme functions?

Question

(define self-add
(let ((x 0))
  (lambda ()
    (set! x (+ x 1))
    x)))

(self-add) => 1

(self-add) => 2

(self-add) => 3

(self-add) => 4

2.

 (define self-add1
    (lambda ()
      (let ((x 0))
        (set! x (+ x 1))
        x)))

(self-add1) => 1

(self-add1) => 1

(self-add1) => 1

Please tell me how to understand the difference between the above two functions? Thanks a lot in advance! Best regards.

Have you tried *using* those functions to see what they do?? — Roddy of the Frozen Peas, Jul 01 '12 at 14:44

Óscar López · Accepted Answer · 2012-07-01T16:24:04.113

5

The first function defines a local variable x with an initial value of 0 and afterwards binds a lambda special form to the name self-add - so x is "enclosed" by the lambda (that's why we say that the lambda from behaves as a closure) and will be the same for all invocations of self-add (you could say that x is "remembered" by self-add), and each time it gets called the value of x will be incremented by one.

The second function binds the lambda to the procedure and afterwards defines a local variable x inside the lambda - here x is redefined each time self-add1 gets called, and will be different for all invocations: so x is never "remembered" by self-add1 and is created anew every time the procedure gets called, initialized with 0 and then incremented, always returning the value 1.

edited Jul 01 '12 at 16:24

answered Jul 01 '12 at 15:21

Óscar López

225,348
35
301
374

Thank you for your detailed explanation! – abelard2008 Jul 01 '12 at 15:58
@abelard20008 My pleasure! welcome to Stack Overflow, please remember to accept the answer that was most helpful for you, by clicking on the check mark to its left. – Óscar López Jul 01 '12 at 16:00
can I think that x in second function is a local variable? thanks – abelard2008 Jul 01 '12 at 16:24
@abelard20008 absolutely, that's what it is: a variable local to each invocation of the function, whereas in the first function `x` is shared by all invocations of the function – Óscar López Jul 01 '12 at 16:28

score 1 · Answer 2 · edited May 23 '17 at 10:09

1

The first function is closure. You create x variable in the lexical scope of the function and the variable holds its value between calls.

Additional information:

edited May 23 '17 at 10:09

Community

1
1

answered Jul 01 '12 at 14:53

Igor Chubin

57,130
8
114
135

where to get the knowledge about Scheme's closure, I didn't find anything about in book "The Scheme programming language 4th edition",Thanks ! – abelard2008 Jul 01 '12 at 15:36

score 0 · Answer 3 · answered Jul 01 '12 at 15:02

0

First function is a closure, while second function simply returns the same function every time, which makes x = 0 and then adds one and returns the result.

answered Jul 01 '12 at 15:02

Mateusz Kowalczyk

2,016
1
14
28

where to get the knowledge about Scheme's closure, I didn't find anything about in book "The Scheme programming language 4th edition",Thanks ! – abelard2008 Jul 01 '12 at 15:29
I recommend that you read [SICP](http://mitpress.mit.edu/sicp/). It's free and a very good Computer Science book that happens to be using Scheme. It covers usage of closures as well. Good luck. – Mateusz Kowalczyk Jul 01 '12 at 17:31

what is the difference between these two Scheme functions?

3 Answers3