On Fri, Sep 6, 2013 at 11:04 AM, Johannes Emerich
Desugaring of an equivalent source file shows that id is applied to the anonymous function, which is then applied to 1.
The following example of a function that is not polymorphic in its return type behaves closer to what I would have expected: It does not work.
Prelude> let z = (\y -> True) :: a -> Bool Prelude> :t (`z` True)
<interactive>:1:2: The operator `z' takes two arguments, but its type `a0 -> Bool' has only one In the expression: (`z` True)
What is the purpose/reason for this behaviour?
Coming from another language, where functions aren't first class, you will probably be used to the notion that a function type is somehow different from any other type. You'll need to unlearn that for functional languages: function types are just as "real" as (Integer) is, and if I have a type variable somewhere which doesn't have constraints otherwise preventing it, that type variable can end up being (Integer) or (a -> a) or (Num c => c -> c -> c) or (Maybe [x]) or (Maybe (a -> a)) or any other (rank-1, i.e. no internal "forall"s) type. (id) has the type (a -> a); in the use mentioned in the first quoted paragraph, this has unified (a) with (b -> b) to produce (id :: (b -> b) -> (b -> b)) in order for the whole expression to be typeable. In your second example, you don't have polymorphism "where it's needed" so it can't infer a type that will work. -- brandon s allbery kf8nh sine nomine associates allbery.b@gmail.com ballbery@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net