Scheme function that returns function is returning unexpected value

Question

I have a function in scheme

    (define (m-m f)
      (let ((count 0))
      (define (h-m-c?) count)
      (define (r-c) (begin (set! count 0) count))
      (define (m-f m)
        (cond ((eq? m 'h-m-c?) (h-m-c))
              ((eq? m 'r-c) (r-c))
              (else (set! count (+ 1 count)) f)))
      m-f))

where m-f is a function that is returned.

However, rather than returning a value, it returns

    #<procedure:m-f>

It appears there is no error with the code, so why doesn't this return a usable value?

What's wrong with a procedure as a value? – dyoo Oct 02 '12 at 05:56 — dyoo, Oct 02 '12 at 05:56

Barmar · Accepted Answer · 2012-10-02T07:17:54.187

You told it to return m-f. That's the name of a procedure that you defined, so it's returning the procedure. To use this return value, you need to assign it to another variable.

(define thing1 (m-m 'unused))
(define thing2 (m-m 'who-cares))
(thing1 'h-m-c?)
(thing1 'increment)
(thing1 'h-m-c?)
(thing2 'h-m-c?)

Some other commentary:

Procedure names ending in ? are used for predicates that report true/false.

You don't need begin in (r-c). Procedure bodies allow multiple expressions, and execute them in order just like begin.

Why didn't you define a function for the case where count is incremented, like you did for the other cases?

Using the default for incrementing seems inappropriate, you should require a specific m parameter for this.

What's the purpose of the f parameter? The only thing it's used for is to be returned when incrementing? Wouldn't the new value of count be a more useful return value? It seems like it's intended to be a "name" for the instance; if so, perhaps you should add a new operation to return the name, rather than doing this when incrementing.

EDIT:

Based on the answer below, I suspect you're misusing the f parameter. It's probably supposed to be a function that you call when doing the increment. So you need to supply it as a lambda expression, and return (f) rather than just f.

Thanks Barmar, I'm new to scheme. I was using the procedure the wrong way. Thanks also for your tips. This procedure counts the number of times another procedure is called, so it takes a procedure as an argument. For example, giving these calls to the m-m:(define m-m (make-monitored (sqrt 100))) ;(m-m 'reset-count) ;(m-m (sqrt 100)) ;(m-m 'how-many-calls?) Again, thanks for your tips! — Matt, Oct 02 '12 at 07:06

Scheme function that returns function is returning unexpected value

1 Answers1