On 10 Dec 2007, at 11:33 AM, Dan Weston wrote:
Questioning apfelmus definitely gives me pause, but...
id :: a -> a -- "arity" 1 id = ($) :: (a -> b) -> (a -> b) -- "arity" 2
I agree with the arities given above (but without quotes) and see no ill-definedness to arity.
But these are two different classes of functions. There are arguments of the first function that cannot be applied to the second (e.g. 5). The fact that the two have different type signatures shows that Haskell can distinguish them (e.g. in the instantiation of a type class)
Not really. The types of id and ($) can't be instances of a type class, since an instance of a type class has to be a monomorphic type. So the decision as to which instance to use has to be made based on the particular monomorphic type id or ($) is used at. But that monomorphic type may still contain free type variables; those type variables themselves represent some single monomorphic type, which may or may not be a function type. So we still don't know what the arity of an arbitrary expression is. (We don't know what its type is, even the way we know the type of id or ($), if it or any of its free variables is lambda-bound).
The difficulties of Haskell's type system in the presence/ intersection of ad hoc/parametric polymorphism is an orthogonal issue. I think that every function application must have a unique monomorphic type at the call site of the "arity" function (assisted or not by type annotation), and this type is known to converge in a Template Haskell construction.
We have to specialize the type of id before supplying it to wrap . For example,
wrap (id :: Int -> Int)
works just fine.
The necessity of type annotation/restriction is an orthogonal issue to the above.
Am I missing something more fundamental?
apfelmus wrote:
Luke Palmer wrote:
Hmm, this still seems ill-defined to me.
compose :: (Int -> Int -> Int) -> (Int -> Int) -> Int -> Int -> Int
Is a valid expression given that definition (with a,b = Int and c = Int -> Int), but now the arity is 4.
:type wrap id wrap id :: (FunWrap (a -> a) y) => [String] -> y but trying to use it like in let x = wrap id ["1"] :: Int yields lots of type errors. We have to specialize the type of id before supplying it to wrap . For example, wrap (id :: Int -> Int) works just fine. I don't like this behavior of wrap since it violates the nice
That's correct, the arity of a function is not well-defined due to polymorphism. The simplest example is probably id :: a -> a -- "arity" 1 id = ($) :: (a -> b) -> (a -> b) -- "arity" 2 Therefore, the polymorphic expression wrap id is problematic. It roughly has the type wrap id ~~ [String] -> a But it's clearly ambiguous: do we have wrap id (x:_) = read x or wrap id (f:x:_) = wrap ($) (f:x:_) = read f (read x) or what? (assuming a read instance for function types) GHCi gives it a type property of polymorphic expressions that it's unimportant when a type variable is instantiated, like in map ((+1) :: Int -> Int) [1..5] = map (+1) ([1..5] :: [Int]) = (map (+1) [1..5]) :: [Int] Regards, apfelmus _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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