"Cristian Baboi" <[EMAIL PROTECTED]> writes: > I think it's a bug. > Here is why: > > let f = (\x -> x/0) in f 0 == f 0 > > Referential transparency say that f 0 must equal to f 0, but in this > case it is not. :-)
I think you are wrong. Referential transparency says that you can replace any occurence of 'f 0' with another expression of the same value, it does not say anything about the behaviour of (==). -k -- If I haven't seen further, it is by standing in the footprints of giants _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
