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
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