The Alternative class says nothing about the a in MyData a. It only
represents code relevant to MyData.
When you see class Applicative f => Alternative (f :: * -> *), that
means all of its functions had to work on any f, where f takes any
type and becomes some other type which could be anything.
The reason it works for Monoid is that class Monoid a where implies
that a is completely known by the time the instance is fulfilled,
therefore the instance can look at what a ended up being and ensure
whatever a is, it must have this constraint on it.
You can tell the difference because mempty returns a type that
mentions the a mentioned in the class, whereas empty returns an a that
is not mentioned in the class, therefore it has to work for any a.
On Wed, Jun 7, 2017 at 8:26 AM, Baa
Hello all!
If I try to write, for example:
instance Alternative MyData where empty = NoMyData a <|> b = if a == b then ...
I get error (see on bottom of the mail) sure, bcz I suppose something like:
Eq a => (MyData a)
All my attempts to add something to instance's `pure` signature have failed. How can I instantiate something with additional restrictions, like in this case? Or are there another solutions for such problem?
Interesting is that MyData derives Eq itself! Which, I suppose, must means that "Eq (MyData a)", and if it's true than "Eq a" is true, because how "MyData a" can be Eq without to be "Eq a" (with *automatically deriving* of Eq instance) ?!
By the way, for Monoid (which is "* -> Constraint") I can add "Eq a" constraint without problems:
instance Eq a => Monoid (Origin a) where mempty = NoMyData mappend NoMyData a = a mappend a NoMyData = a mappend (MyData a) (MyData b)|a == b = MyData a |otherwise = NoMyData
but not for Alternative, which is "(* -> *) -> Constraint".
*ORIGINAL ERROR DUMP*: ====================== 42 16 error error: • No instance for (Eq a) arising from a use of ‘==’ Possible fix: add (Eq a) to the context of the type signature for: (<|>) :: Origin a -> Origin a -> Origin a • In the expression: a == b In the expression: if a == b then NoOrigin else NoOrigin In an equation for ‘<|>’: a <|> b = if a == b then NoOrigin else NoOrigin (intero)
/Best regards Paul _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners