If you don't want to use the package or you want other criteria for
stooping, I think the correct way to do this are continuation marks, but an
easier but more inefficient way is using a parameter (so it is thread safe).
;-----
#lang racket
(define recursion-level (make-parameter 0))
(define (f x)
(sleep (* .1 (random)))
(cond
[(<= (recursion-level) 5)
(parameterize ([recursion-level (add1 (recursion-level))])
(displayln (list (recursion-level) x))
(f (* x 2)))]
[else
(displayln (list "too many recursions" x))]))
(thread (lambda () (f 10)))
(f 1)
; -----
Gustavo
On Fri, Aug 9, 2019 at 2:03 PM Sam Tobin-Hochstadt <[email protected]>
wrote:
> This is possible -- the central trick is to maintain information about
> the call stack separately, perhaps by using continuation marks.
>
> We (mostly Phil Nguyen) recently built a tool that does this. If you
> take your program, install the "termination" package, and then add:
>
> (require termination)
> (begin/termination (recur1 0))
>
> you get this error message:
>
> possible-non-termination: Recursive call to `#<procedure:recur1>` has
> no obvious descent on any argument
> - Preceding call:
> * 1st arg: 3
> - Subsequent call:
> * 1st arg: 6
>
> We have a paper, here: https://arxiv.org/abs/1808.02101 which goes
> into detail about how it works. The source code is here:
> https://github.com/philnguyen/termination
>
> Sam
>
> On Fri, Aug 9, 2019 at 11:58 AM Kees-Jochem Wehrmeijer
> <[email protected]> wrote:
> >
> > Hi,
> >
> > I'm wondering whether it's possible for a function to detect whether
> it's being called by itself, possibly indirectly. I want to use this to
> prevent endless loops and generate different results in these situations.
> For example, imagine I have the following functions:
> >
> > (define (foo loc)
> > (list
> > (cons loc 'a)
> > (cons (+ loc 1) 'b)
> > (cons (+ loc 2) 'c)))
> >
> > (define (combine a b)
> > (lambda (loc)
> > (append (a loc) (b (+ loc 3)))))
> >
> > So I can do:
> > > ((combine foo foo) 0)
> > => '((0 . a) (1 . b) (2 . c) (3 . a) (4 . b) (5 . c))
> >
> > Now, let's say I have these recursive definitions:
> >
> > (define (recur1 loc)
> > ((combine foo recur2) loc))
> >
> > (define (recur2 loc)
> > (recur1 loc))
> >
> > Defined in this way, a call to (recur1 0) loops endlessly. I would like
> it too return something like
> > > (recur1 0)
> > => '(0 . a) (1 . b) (2 . c) (loop 0))
> >
> > So I was thinking if I could make it work if I could do something like:
> > (define (recur1 loc)
> > (if (in-recursive-call?)
> > (list (cons 'loop 0))
> > ((combine foo recur2) loc)))
> >
> > Is that something that's possible? Is there a better way to do what I
> want?
> >
> > Thanks,
> > Kees
> >
> > --
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