[Haskell-cafe] Re: Justification for Ord inheriting from Eq?
maeder at tzi.de
Fri Apr 7 04:58:23 EDT 2006
John Meacham wrote:
> 1. one really does logically derive from the other, Eq and Ord are like
> this, the rules of Eq says it must be an equivalance relation and that
> Ord defines a total order over that equivalance relation. this is a good
> thing, as it lets you write code that depends on these properties.
given an Ord instance (for a type T) a corresponding Eq instance can be
instance Eq T where
a == b = compare a b == EQ
This does not make the definition of an Ord instance (that is supposed
to match an equivalence) easier but ensures at least the required
I never dared to define this generically and switch on various ghc
instance Ord a => Eq a where ...
It just strikes me that wrt. "NaN" the Ord instances for Float and
Double are no total orders (and thus these types are not suited for
sets, i.e. "Data.Set.Set Float")
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