On Fri, Oct 29, 2010 at 8:33 AM, Tillmann Rendel
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