[Haskell-beginners] how do typeclasses work again?

[David McBride]

foo2 :: (Is isx Integer) => isx -> String


isx -> String - That means that this function takes anything and
returns a string.
Is isx Integer => - That just means that whatever isx is, there should
be an Is isx Integer instance that satisfies it.

Putting those together this function takes anything and returns a
string, so long as the anything (isx) satisfies the constraint I isx
Integer.

But there's nothing in the type or code that says what type x actually
is.  The Integer in the constraint just constrains what isx can be.

To fix it add the ScopedTypeVariables extension and try this:

foo2 :: (Is isx Integer) => isx -> String
foo2 = apply (\(i :: Integer) -> "")

Alternatively if you are using ghc 8, you can turn on TypeApplications
and use this:

foo2 :: (Is isx Integer) => isx -> String
foo2 = apply @_ @Integer (\i -> "")

On Thu, Feb 9, 2017 at 11:59 AM, Nicholls, Mark <nicholls.mark at vimn.com> wrote:
>
> Sorry..I do haskell about once every 6 months for 2 hours...and then get on with my life.
>
> I always forget some nuance of typeclasses.
>
> Consider some simple typeclass
>
>> class Is isx x where
>>   apply :: (x -> y) -> isx -> y
>
>
> We can make any type a member of it...mapping to itself
>
>> instance Is x x where
>>   apply f = f
>
> But we can also make a tuple a member of it...and pull the 1st member..
>
>> instance Is (x,y) x where
>>   apply f (x,y) = f x
>
> Weird and largey useless...but I'm playing.
>
> Then construct a function to operate on it
>
>> foo2 :: (Is isx Integer) => isx -> String
>> foo2 = apply (\i -> "")
>
> And...
>
> • Could not deduce (Is isx x0) arising from a use of ‘apply’
>       from the context: Is isx Integer
>         bound by the type signature for:
>                    foo2 :: Is isx Integer => isx -> String
>         at prop.lhs:51:3-43
>       The type variable ‘x0’ is ambiguous
>       Relevant bindings include
>         foo2 :: isx -> String (bound at prop.lhs:52:3)
>       These potential instances exist:
>         instance Is x x -- Defined at prop.lhs:41:12
>         instance Is (x, y) x -- Defined at prop.lhs:45:12
>     • In the expression: apply (\ i -> "")
>       In an equation for ‘foo2’: foo2 = apply (\ i -> "")
>
>
> What's it going on about?
> (my brain is locked in F# OO type mode)
>
> I've told it to expect a function "Integer -> String"...surely?
> Whats the problem.
>
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