[r6rs-discuss] [Formal] eliminate library export immutability loophole

From: AndrevanTonder <andre>
Date: Mon Mar 12 18:05:47 2007

On Sun, 11 Mar 2007, R. Kent Dybvig wrote:

> Okay, here's one example:
>
> (library (L1)
> (export y get-y set-y!)
> (import (r6rs))
> (define x (call/cc (lambda (k) (list 0 k values))))
> (define y (car x))
> (define z ((caddr x)))
> (define get-y (lambda () y))
> (define set-y!
> (lambda (v)
> (call/cc (lambda (k) ((cadr x) (list v (cadr x) k)))))))

I think there is an error. Here is a working version:

    (library (L1)
      (export y get-y set-y!)
      (import (r6rs))
      (define x (call/cc (lambda (k) (list 0 k values))))
      (define y (car x))
      (define z ((caddr x) #f)) ; modified line
      (define get-y (lambda () y))
      (define set-y!
        (lambda (v)
          (call/cc (lambda (k) ((cadr x) (list v (cadr x) k)))))))

> By my reading of the current library description, the program:
>
> (import (r6rs) (L1))
> (write (list y (get-y))) (newline)
> (set-y! 3)
> (write (list y (get-y))) (newline)
>
> prints
>
> (0 0)
> (0 3)

I get the following, which I think is also consistent with the current
specification:

   (0 0)
   (3 3)

This happens if exported variables are translated to shared r5rs toplevel
globals, which is a possible implementation of the shared semantics.

Andre
Received on Mon Mar 12 2007 - 18:05:10 UTC