[Haskell-cafe] Re: type class question

Peter Padawitz peter.padawitz at udo.edu
Mon Dec 10 10:12:01 EST 2007

Jules Bean wrote:

> Peter Padawitz wrote:
>> Jules Bean wrote:
>>> Peter Padawitz wrote:
>>>> What is so bad about making compFoo part of the class? It reduces 
>>>> the code (constraints can be avoided) and reflects the close 
>>>> connection between  a signature Sig (implemented by the class) and 
>>>> the evaluation (compFoo) of Sig-terms in Sig-algebras.
>>> making it part of the class allows instances to override the 
>>> implementation.
>>> Which in this case is a strange thing to do.
>> Sure, but this can only happen because Haskell does not check whether 
>> the instances satisfy the equations in the class. The type class 
>> concept would be "cleaner" if all methods (partially or totally) 
>> defined by equations within the class were not allowed to be 
>> instantiated!
> I don't see why!
> In the class
> class Foo a where
>   f :: a -> Int
>   g :: b -> Integer
>   g = fromIntegral . f
> The equations within the class are defaults, not equations. 

I must admit that I didn't know this... Nevertheless, won't you agree 
that the default and the actual instance should be semantically equivalent?

> The equation for 'g' is a default, not a rule.
> If you want equations, you do it outside the class. I have written 
> that class wrongly, I should actually write g = fromIntegral . f as a 
> function outside the class, thus guaranteeing the implementation and 
> stopping people breaking that invariant.
> The purpose of methods with defaults is to allow the possibility that 
> there is an obvious natural way to implement one function in terms of 
> others, but there might be more efficient ways.
> For example, the Foldable class should (but doesn't) have a member 
> length. This could be defaulted to length . toList, but have a more 
> efficient implementation in Sequence, which stores its own length anyway.
> Or maybe we are at cross-purposes.

No no, default functions make sense.

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