[R6RS] non-opaque record equality

William D Clinger will at ccs.neu.edu
Sat Feb 25 16:43:29 EST 2006

Kent wrote:
>   (equal? (f) (f))      nonterminating if procedures are records,
>                         #f otherwise

If procedures were records, wouldn't they be *opaque*
records?  In which case the recursive comparison would
be eqv?, and the equality comparison would terminate.

Another way of saying it is that representing procedures
as non-opaque records would be illegal if equal? compares
all such records recursively, because the R5RS requires
equal? to use eqv? when comparing procedures.

I'm not a big fan of defining equal? to compare records
recursively, and I'm on record as opposing this in the
past, but having it do so on *non-opaque* records but
not on *opaque* records seems like a reasonable and
intuitive compromise.

Another possibility is that we could require equal? to
terminate on all arguments.  Judging by a discussion in
comp.lang.scheme, it appears this is easier and more
efficient than I had  believed when we last had this


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