1. I don't understand why you decrement the count in operator position
2. I don't see that you increase the count when a procedure is returned
from a lambda

Example
(define (f a) a)
(define (g)
    (hash-set! H f 1) ; (*)
    (f 1)) ;(**)
(define (h)
   (pk (hash-ref H f))) ; (*)

(g)
(h)

=> '(#f) as count in your patch is counted up twice (*) and down ones (**)
in total count = 1 so you will not maintain the identity of f and you will
get a bad printout

Then we also have this example
(define (f a) a)
(define (u) f)
(define (g) (hash-set! H (u) 1))
(define (h) (pk (hash-ref H f)))

(g)
(h)
This will again print (#f) as the count will be 1.

/Stefan























On Tue, Jan 14, 2020 at 5:16 PM Andy Wingo <wingo@pobox.com> wrote:

> On Tue 14 Jan 2020 15:47, Stefan Israelsson Tampe <stefan.itampe@gmail.com>
> writes:
>
> > Yes, your patch is indicating when you should use the same identity
> > e.g. all uses of procedures in a higher order position such as an
> > argument or a return value. But I looked at your patch, which looks
> > good but I saw that for operator position you decrease the count. Why?
> > Also you are free to use one version in argument / return position and
> > another one in operator position the only limit is to use the same
> > identity for on operator position. Finally don't you need to count
> > usage of returning a variable as well?
>
> Not sure what the bug is.  Do you have a test case that shows the
> behavior that you think is not good?
>
> Andy
>