[r6rs-discuss] Why is (eqv? g g) unspecified when g is a procedure?
bear at sonic.net
Mon Jul 16 10:29:39 EDT 2012
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On 07/16/2012 05:37 AM, William D Clinger wrote:
> John Cowan wrote:
>> I'm probably missing something here, but if h doesn't need a
>> closure, what's wrong with using the code address as a unique
> Because that would violate both R5RS and R6RS semantics.
> 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.
Um? Will, did you misspeak here or is there something very
fundamental that I'm not understanding? As I see it the same
vector of code ought to produce the same behavior when run.
Different nonempty closures over the same procedure are unique,
of course, but the procedure itself?
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