On Fri, Oct 29, 2010 at 8:33 AM, Tillmann Rendel <[email protected]> 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 (==)? -- Dave Menendez <[email protected]> <http://www.eyrie.org/~zednenem/> _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
