[Haskell-beginners] typeclass woes...how to constain a typeclass to be "closed" under an operation....

Wed Jan 7 13:20:17 UTC 2015

Ouch I’m now posting about the same problem in both Haskell mailing lists….

The problem was that if you comment out

--instance Bar (S Integer)

The code still compiles…the “Bar (S a)” isnt evaluated until someone invokes inc’….I want my operation to be closed by definition….in advance.

A solution has been posted in the café, which is basically the same as the one I posted here, but fixes the cyclic typeclass error..

> class Foo_ a -- just to prevent a cycle in superclass constraints

> instance Foo a => Foo_ a

> class Foo_ (S a) => Foo a where
>   type S a
>   op :: a -> (S a)

> -- and an example where you get a compile error if "op x" has an instance,
> -- but "op (op x)" does not have an instance.
> instance Foo Int where
>   type S Int = Char
>   op = toEnum

> instance Foo Char where
>   type S Char = (Char,Char)
>   op x = (x,x)

> instance Foo (Char,Char) where
>   type S (Char,Char) = Int
>   op (x,y) = fromEnum x

I’ve now tried to write some code using both “idioms”….i.e. the above…and the

The above idiom forces you to define an associated type that inhabits  the typeclass

The *->* (e.g. monad) idiom forces you to define a parameter to infer the return type, that also inhabits the typeclass…(technically I suppose “m b” doesn’t inhabit the typeclass, “m” does….but

So…

(>>=)<http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:-62--62--61-> :: Monad m => m a -> (a -> m b) -> m b<http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:-62--62--61->

Is closed.

(S Integer) is not an instance of Bar.  I solved this by doing the following
You can get around this using an empty instance declaration for (S Integer) like so
instance Bar (S Integer)

Now, the compiler will know that (S Integer) is an instance of this.  It doesn't matter if there is no definition of inc' as you won't be using it.
However, I don't understand why a class definition will contain itself as a constraint on one of its method.  By doing so,

That is, whenever you do something like this

class Bar a where
  inc' ::  Bar (S a) => a -> S a

instance Bar Integer where
  inc' x = S x
The resulting expression will also have to be an instance of Bar (.ie.,instance Bar (S Integer)
)  I just don't think this makes any sense.  Also, tried searching for any such type signatures in Hoogle, but didn't come across anything like this.

On Tue, Jan 6, 2015 at 11:09 PM, Nicholls, Mark <nicholls.mark at vimn.com<mailto:nicholls.mark at vimn.com>> wrote:
I can do the basic Haskell…I’ve done a course even, but as I don’t use it… I quickly degenerate into a beginner.

I think I’ve refound my feet….I will post it to the café and see what happens.

From: Beginners [mailto:beginners-bounces at haskell.org<mailto:beginners-bounces at haskell.org>] On Behalf Of David McBride
Sent: 06 January 2015 4:41 PM
To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell
Subject: Re: [Haskell-beginners] typeclass woes...how to constain a typeclass to be "closed" under an operation....

I think you've slightly surpassed the skill level of this list.  You might try on haskell-cafe, and I would bet you could get an in depth answer on stackoverflow.

On Tue, Jan 6, 2015 at 3:46 AM, Nicholls, Mark <nicholls.mark at vimn.com<mailto:nicholls.mark at vimn.com>> wrote:
I dont seem to have any replies
Booo
Is my question too stupid? Or boring? Or unintelligable?

Its quite common in maths to have operations in a theory that are (set) closed, i just want to translate that notion to a typeclass

I have a suggestion that does work

Class Foo m where
   op :: m a -> m (S a)

That is closed, but now were working on types of kind • -> •, which is another leap of complexity

Is this the idiom/pattern i should follow? Or can the closure contraint be expressed directly?

Excuse the spelling, sent from a phone with itty bitty keys, it like trying to sow a button on a shirt with a sausage.

On 5 Jan 2015, at 10:54, Nicholls, Mark <nicholls.mark at vimn.com<mailto:nicholls.mark at vimn.com>> wrote:
> {-# LANGUAGE TypeFamilies, MultiParamTypeClasses, FlexibleContexts,  FlexibleInstances, GADTs #-}

lets say I have a data type

> data S a = S a deriving Show

and a nice simple typeclass

> class Foo a where
>   inc :: a -> S a

make Integer an instance and....

> instance Foo Integer where
>   inc x = S x

brilliant...

now lets say I want my typeclass to enforce that any result of inc is also of a type that is an instance of the typeclass
I'll follow my nose...which causes me a problem that makes me realise I maybe I don't understand whats going on.

> class Bar a where
>   inc' :: (Bar (S a)) => a -> S a

follow it to....

> instance Bar Integer where
>   inc' x = S x

this WORKS!...even though "S Integer" is not an instance of Bar!...maybe I don't understand.

> -- x = inc' (1 :: Integer)

This would go BOOM ....”no instance of (Bar (S Integer))"
...so that makes sense....

How do I force Haskell to check the constraint before I evaluate an expression?

follow my nose?

> -- class (Baz (S a)) => Baz a where
> --  inc'' :: a -> S a

BOOM ...."Cycle in class declaration"....
booo..but I suppose thats understandable....

so

a) is my use of typeclasses correct?...or am I missing something?
b) if so, how do I force the compiler to make sure my typeclass operations is nicely closed (under the typeclass)?

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