[r6rs-discuss] Why is (eqv? g g) unspecified when g is a procedure?
cowan at mercury.ccil.org
Mon Jul 16 12:18:44 EDT 2012
Aaron W. Hsu scripsit:
> William D Clinger <will at ccs.neu.edu> wrote:
> > Both the R5RS and R6RS insist that procedures that are eqv?
> > behave the same. Distinct procedures that share the same
> > code are unlikely to behave the same.
> This is true in general, but I think the assertion was that two
> procedures that do not need closures to run correctly, can only be
> observationally different or distinct if they have different code bodies.
> I am not seeing a fault in this reasoning off the top of my head.
That's what I meant, allowing users to reliably get (eqv? eqv? eqv?) => #t.
Real FORTRAN programmers can program FORTRAN John Cowan
in any language. --Ed Post cowan at ccil.org
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