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From: gyro@cymbal.reasoning.COM (Scott Layson Burson)
Newsgroups: comp.lang.scheme
Subject: C->Scheme mappings
Message-ID: <9105302308.AA18221@cymbal.reasoning.com.>
Date: 30 May 91 23:08:38 GMT
References: <1991May28.191834.15025@apd.mentorg.com>
Sender: daemon@athena.mit.edu (Mr Background)
Reply-To: Gyro@reasoning.com
Organization: The Internet
Lines: 54


   Date: 28 May 91 19:18:34 GMT
   From: "Joe Mueller @ APD x1367" <joemu@mentorg.com>

   I'm trying to map the various C control statements into
   Scheme. The translation of break, continue, and return
   are giving me trouble. (I am not a scheme programmer so
   an obvious answer may not be obvious to me)

   I suspect I need to do something with
   call-with-current-continuation but I don't see exactly how
   to do it. Anyone care to help?

You don't need call-with-current-continuation.

`return' is implicit in Scheme: a lambda body always returns the value of its
last form.  So you don't have to do anything special with it; e.g.

   int foo() { return 3; }
  -->
   (define (foo) 3)

The way `break' and `continue' work is tied up in the way iteration is modelled
using tail-recursion.  For instance

   for (i = 0; i < 10; ++i) {
        if (p1(i)) continue;
        /* ... <1> */
        if (p2(i)) break;
        /* ... <2> */
   }

 -->

   (letrec (((loop i)
               (if (not (< i 10)) #f
                 (if (p1 i) (loop (+ 1 i)))  ; `continue'
                 ;; ... <1>
                 (if (p2 i) #f   ; `break' value necessary
                   (begin
                      ;; ... <2>
                      (loop (+ 1 i)))))))
      (loop 0))

(Apologies if I screwed up the syntax -- I haven't written much Scheme lately.)

Here you see that the `continue' has turned into `(loop (+ 1 i))', which
invokes the next cycle of the loop, and `break' has turned into `#f' (it
doesn't matter what the value is, but there has to be one).  That is, `break'
amounts to simply not invoking the next iteration.

A bit mind-bending, eh?

-- Scott