Re: [Haskell-beginners] Making a generic interpreter

On Saturday 07 May 2011 11:39:44, Erik Helin wrote:

Thank you Patrick for posting your code, it helps a lot too see a complete example!

On Sat, May 7, 2011 at 03:11, Patrick LeBoutillier

wrote:

...

instance AbsInteger Integer where a `add` b = (+) a b a `eq` b = absBool $ a == b absInteger a = a

In my code, I did not use "a `eq` b = absBool $ a == b". Instead, I used "a `eq` b = a == b". I thought, since "a == b" is of type Bool, and Bool is an instance of AbsBool, I can just return type Bool.

Why isn't this the case? What is it about typeclasses that I'm not understanding?

A type signature foo :: (Bar b) => SomeType -> b says "given a value of SomeType, I can produce return values of *any* type, as long as it's an instance of Bar". The caller decides which type it wants and the callee must be able to produce it. In OO, on the other hand, the callee decides which type it returns, a signature Bar foo(SomeType s); (where Bar is an interface) says foo will return values of a type, all you know about which is that it implements Bar. Using explicit quantification, the Haskell type is foo :: forall b. (Bar b) => SomeType -> b while the OO-type would correspond to foo :: exists b. (Bar b) => SomeType -> b (the latter type doesn't exist in Haskell).