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

[Aaron W. Hsu]

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.

-- 
Aaron W. Hsu | arcfide at sacrideo.us | http://www.sacrideo.us
Programming is just another word for the lost art of thinking.