Re: [Scheme-reports] Procedural equivalence: the last debate

[Alexey Radul <axch@x>]

Works for me, as would an implementation of Shinn's proposal.

~Alexey

On Sun, Jun 9, 2013 at 2:00 AM, Arthur A. Gleckler
<scheme@x> wrote:
> On Sat, Jun 8, 2013 at 10:48 PM, John Cowan <cowan@x> wrote:
>
>>
>> +1, with the addition of records after pairs and bytevectors after
>> vectors.
>
>
> Yes, sorry, I should have made that adjustment.  I hereby revise my
> proposal.  Below is the language after your revision, starting from the
> phrasing from R7RS draft 9:
>
>   The eqv? procedure returns #t if... obj1 and obj2 are procedures whose
>   location tags are equal.
>
>   The eqv? procedure returns #f if... obj1 and obj2 are procedures
>   that would behave differently (return different value(s) or have
>   different side effects) for some arguments.
>
>   The above definition of eqv? allows implementations latitude in
>   their treatment of procedures and literals: implementations are free
>   either to detect or to fail to detect that two procedures or two
>   literals are equivalent to each other, and can decide whether or not
>   to merge representations of equivalent objects by using the same
>   pointer or bit pattern to represent both.
>
>   On symbols, booleans, the empty list, pairs, procedures, records,
>   non-empty strings, vectors, and bytevectors, eq? and eqv? are
>   guaranteed to have the same behavior.
>
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