Re: [Haskell-beginners] how do typeclasses work again?

lovely so if I now go....

foo4 = apply (\i -> show i)

And :t foo4 is...

foo4 :: (Is isx a, Show a) => isx -> String

So add that as a type...and we get the same sort of awfulness...."could not deduce" bla bla But how do you make this disappear

foo4 = apply (\(i :: a) -> show i)

Doesn’t work... "could not deduce" bla bla I'd instinctively like to go....

foo4 = apply (\(i :: ((Show a) => a)) -> show i)

"Illegal qualified type: Show a => a" And this is really just

foo4 = apply show

Where we end with "could not deduce" Sorry....I'm struggling.

-----Original Message----- From: Beginners [mailto:beginners-bounces@haskell.org] On Behalf Of David McBride Sent: 09 February 2017 5:31 PM To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Subject: Re: [Haskell-beginners] how do typeclasses work again?

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 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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