David (cc'ing larceny-use4rs)- On Fri, Apr 17, 2009 at 3:17 PM, David Rush <[email protected]> wrote: > I suspect that this is a behavior specified by R6RS, but why isn't > +inf.0 EQ? to itself? It is currently EQV? and EQUAL?, so I'm a little > confused.
The *value* produced from evaluating the expression +inf.0 actually is eq to "itself," as one would expect (for any object), as illustrated below. > (eq? +inf.0 +inf.0) #f > (let ((x +inf.0)) (eq? x x)) #t > (inexact? +inf.0) #t Beyond that, I do not understand your question; why would any two boxed objects necessarily be eq? (All inexacts in the current version of Larceny are boxed.) -Felix _______________________________________________ Larceny-users mailing list [email protected] https://lists.ccs.neu.edu/bin/listinfo/larceny-users
