On Tue, Feb 07, 2006 at 07:59:46PM -0800, Ashley Yakeley wrote:
John Meacham wrote:
however, (Set (a -> a)) is malformed. since a _requirement_ is that Set can only be applied to a type with an Eq constraint so the instance you try to do something like
returnid :: Set (a -> a) -- ^ static error! you need
returnid :: Eq (a -> a) => Set (a -> a)
the instant you introduce 'Set' you introduce the 'Eq' constraint. as long as you are just working on generic monads then there is no need for the Eq constraint.
OK, try this:
foo :: (Monad m) => m Int foo = return id >>= (\i -> i 7)
fooSet :: Set Int fooSet = foo
Since we have (Eq Int), your type-checker should allow this. But your instance implementation of return and (>>=) made assumptions about their arguments that foo does not stick to.
I assume you mean:
foo = return id >>= (\i -> return (i 7))
Yup. and this is just fine since Int is an instance of Eq. this should typecheck and does. foo is a completly generic function on any monad, the particular instance for 'Set' places the extra restriction that sets argument must be Eq. this need not be expressed in foo's type because it is polymorphic over all monads, however as soon as you instantiate foo to the concrete type of 'Set a' for any a, the Eq constraint is checked and in the dictionary passing scheme, the appropriate dictionary is constructed and passed to foo. it is important to realize that dictionaries are unchanged with this translation. a Monad dictionary is a monad dictionary whether it depends on an Eq instance (because it is an instance of a restricted data type) or not. 'foo' expects a monad dictionary just like any other, that said dictionary is created partially from Ints Eq instance is immaterial. John -- John Meacham - ⑆repetae.net⑆john⑈