I'll apply your patch and see if it works. After reading it more carefully
I think I understand your decrement count. Nice code!

On Tue, Jan 14, 2020 at 5:36 PM Stefan Israelsson Tampe <
stefan.itampe@gmail.com> wrote:

> 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
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> 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
>>
>