John Cowan <[email protected]> writes: > Taylan Ulrich B. scripsit: > >> eq? is a (typically) more efficient eqv?, > > I have yet to be convinced that `eq?` actually is more efficient > than `eqv?` on the safe types.
I suppose type-inference can reduce many uses of eqv? to eq?, and when inference fails, the difference is merely a dispatch on the type-tag. It still cannot be as efficient in absolutely all cases though, can it? Perhaps the total cost of globally substituting eqv? for eq? is not as big as I would've imagined, given a sufficiently intelligent compiler. _______________________________________________ Scheme-reports mailing list [email protected] http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports
