Re: [Haskell-cafe] Questions about the Functor class and it's use in "Data types à la carte"

[Roberto Zunino]

Dominic Steinitz wrote:
> Roberto Zunino wrote:
>>This is the point: eta does not hold if seq exists.
>>
>>  undefined `seq` 1 == undefined
>>  (\x -> undefined x) `seq` 1 == 1
> 
> Ok I've never used seq and I've never used unsavePerformIO. Provided my
> program doesn't contain these then can I assume that eta reduction holds
> and that (.) is categorical composition?

Yes, provided that you do not use seq and all its related stuff, e.g. 
($!), foldl', bang patterns, data Foo a = Foo !a, ...

Also, note that you still can define and use seq restricted to many 
useful types:

    seqInt :: Int -> a -> a
    seqInt 0 x = x
    seqInt _ x = x

IIRC, you can also have a quite general

    seqData :: Data a => a -> b -> b

>>The "(.) does not form a category" argument should be something like:
>>
>>  id . undefined == (\x -> id (undefined x)) /= undefined
>>
>>where the last inequation is due to the presence of seq. That is,
>>without seq, there is no way to distinguish between undefined and (const
>>undefined), so you could use a semantic domain where they coincide. In
>>that case, eta does hold.
> 
> It would be a pretty odd semantic domain where 1 == undefined. Or
> perhaps, I should say not a very useful one.

in the new domain, you do not have 1 == undefined (which are still 
different) but merely

   undefined == (\x -> undefined)

Regards,
Zun.