[r6rs-discuss] Why is (eqv? g g) unspecified when g is a procedure?

[John Cowan]

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