Re: [Scheme-reports] eq? and eqv? for records

On Fri, Feb 14, 2014 at 10:17 PM, Taylan Ulrich Bayırlı/Kammer <taylanbayirli@x> wrote:

The `eqv?' definition says that records are equivalent if denoting the
same location and points to section 3.4, which explains that the notion
of "storage being newly allocated" is what denotes the creation of
objects with distinct locations, yet section 5.5 (<constructor name>
point) doesn't use that phrase. In short, we base record equivalence
semantics on their location, yet don't specify their location.

Correct.

The section needs to be expanded to specify at least the discerning of
obviously inequivalent records (mutable and created with distinct calls
to the constructor, or immutable but having non-eqv? fields). (Note
that we already support immutable record types by omitting mutators.)

There are two issues, of course - specifying under what

circumstances, if any, a record constructor must return

a newly allocated record, and specifying the eqv? semantics.

"The returned record must be newly allocated if any <modifier name> is
provided, or if not all fields have an initial value that is equivalent
per eqv? (section 6.1) to the corresponding field value of a record of
the same type."

I'm not sure what you're trying to specify about the

initial values here, but it may be simpler to just leave

this out. "The constructor guarantees a newly allocated

record if there are any mutable fields." You can't assume

anything about immutable records, including the degenerate

empty record case.

[Note, simply not providing a <modifier> is not sufficient

to guarantee immutability in the presence of SRFI 99 style

record introspection.]

Ideally, IMO, the equivalence semantics of all objects created with
constructors would follow the logic we used for procedures

Ah, well this was rather controversial itself :) While

a stronger case can be made for desiring predictable

record identity, similar optimizations still apply. This

isn't really something we can just write in as an errata.

Maybe in R8RS. :-)

Yes, I think so.

Alex