Hi, sorry for answering to such an old thread.
David Menendez <d...@zednenem.com> writes: > On Fri, Oct 29, 2010 at 8:33 AM, Tillmann Rendel > <ren...@informatik.uni-marburg.de> wrote: >> Hi, >> >> Uwe Schmidt wrote: >>> >>> In the standard Haskell classes we can find both cases, >>> even within a single class. >>> >>> Eq with (==) as f and (/=) as g belongs to the 1. case >> >> Note that the case of (==) and (/=) is slightly different, because not only >> can (/=) be defined in terms (==), but also the other way around. The >> default definitions of (==) and (/=) are mutually recursive, and trivially >> nonterminating. This leaves the choice to the instance writer to either >> implement (==) or (/=). Or, for performance reasons, both. > > While I understand the argument in general, I've never understood why > it applies to Eq. Are there any types where it is preferable to define > (/=) instead of (==)? Yes for infinite data structures. Cheers, Jean-Marie _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
