G'day all. Quoting Janis Voigtlaender <[EMAIL PROTECTED]>:
Yes. But actually what we would need would be that it checks as well that we have implemented at *most* a minimal set of operations. Otherwise, we are back to the point where I can implement both (==) and (/=), and in a way that the supposed invariant is broken.
Except that there are many circumstances where I can write an operation that's more efficient (or more lazy, or whatever) than the default, even though they do the same thing. Probably not true of (==) vs (/=), but it's often true of (<=) vs compare. Cheers, Andrew Bromage _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
