[Haskell-cafe] Why can't we make an instance declaration on a
brandon.m.simmons at gmail.com
Mon Jan 4 12:57:38 EST 2010
On Mon, Jan 4, 2010 at 12:11 PM, Reid Barton <rwbarton at math.harvard.edu> wrote:
> On Mon, Jan 04, 2010 at 11:50:57AM -0500, Brandon Simmons wrote:
>> On Sun, Jan 3, 2010 at 2:22 AM, Ivan Lazar Miljenovic
>> <ivan.miljenovic at gmail.com> wrote:
>> > jberryman <brandon.m.simmons at gmail.com> writes:
>> >> This may be a dumb question, but why can we not declare a Monad
>> >> instance of a type synonym? This question came to me while working
>> >> with the State monad recently and feeling that the requirement that we
>> >> wrap our functions in the State constructor is a bit... kludgy.
>> > Because type defines an _alias_. If you define "type Foo = Maybe Int",
>> > then everywhere you have a "Foo" the compiler should be able to replace
>> > it with "Maybe Int".
>> > As such, if you have a custom instance on your type synonym (say a
>> > custom Show instance for Foo), then which instance will the compiler
>> > use?
>> Thanks. I guess what I'm really asking is if there is any way to
>> redefine the monad instance for (->) such that we can have a State
>> monad without the data constructor wrapper.
> It would be a nightmare for type inference. Consider:
> return "foo" :: String -> (String, String) -- \x -> ("foo", x)
> which would be valid in a language where we can do what you suggest.
> `return` has type `a -> m a`, so `a` must be `String`, and now we need
> to unify `String -> m String` with `String -> String -> (String, String)`.
> In Haskell, these just don't unify because there are no type-level lambdas.
> Even if there were, how is the typechecker supposed to know that we want
> the solution `m a = String -> (a, String)` and not `m a = a -> (a, a)`
> or many other possibilites?
> The purpose of the newtype State is to have something to unify `m` with:
> return "foo" :: State String String
> We can unify `String -> m String` with `String -> State String String`
> by setting `m` to `State String`.
> Reid Barton
Thanks, that makes a lot of sense.
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