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

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.