26 Apr
2010
26 Apr
'10
2:55 p.m.
On 26-4-2010 20:12, Daniel Fischer wrote:
Am Montag 26 April 2010 19:52:23 schrieb Thomas van Noort:
...
Yes, y's type is more general than the type required by f, hence y is an acceptable argument for f - even z :: forall a b. a -> b -> Bool is.
That's what I thought. I've just never seen such a notion of a more general type involving overloading before.
However, it requires y to throw away the provided dictionary under the hood, which seems counter intuitive to me.
Why? y doesn't need the dictionary, so it just ignores it.
Sure, but y's type explicitly mentions that it doesn't want a dictionary, so why would you provide one to it?
Regards, Thomas
Regards, Thomas