How overload operator in Haskell?

[Dylan Thurston]

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On Fri, Jul 11, 2003 at 05:38:18PM +1000, Andrew J Bromage wrote:
> G'day all.
>=20
> On Thu, Jul 10, 2003 at 11:16:56PM -0700, Ashley Yakeley wrote:
>=20
> > As written, this is _not_ a good idea. Trust me, you end up having to=
=20
> > put type annotations everywhere. Even (3 + 4 :: Integer) is ambiguous,=
=20
> > you have to write (3 :: Integer) + (4 :: Integer).
>=20
> But that's what default() is for!

Don't be silly: the default mechanism is a very special case that is not
good for general use.  It would make all user-defined numeric types
essentially unuseable, for one thing.

(Unless you have a proposal to make the default mechanism more general?)

Peace,
	Dylan

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