I've started Haskell not too long ago, and am now teaching myself about monads, so I've come up with a dumb example for myself here to assist me in understanding an aspect of monads.
Given a type Certainty defined as
data Certainty a = Definitely a | Perhaps a | Nope
and an instance of Monad for said type
instance Monad Certainty where
return = ...
how would the method return be defined for Certainty and each of its data constructors?
you have to pick one - based on the naming I would suggest
return value = Definitely value
remember that return a >>= f must be f a
bind should probably be
(Definitely a) >>= f = f a
(Perhaps a) >>= f = case f a of
Definitely b -> Perhaps b
Perhaps b -> Perhaps b
Nope -> Nope
Nope >>= _ = Nope
Carsten's answer is quite correct, of course, but it is, in my experience, often more intuitive to specify a monad by defining the join function (especially in this case, where any child will understand that if Mum says that we will perhaps definitely have ice cream tomorrow that means that we will perhaps have ice cream..)
join :: Certainty (Certainty a) -> Certainty a
join Nope = Nope
join (Perhaps Nope) = Nope
join (Perhaps (Perhaps x)) = Perhaps x
join (Perhaps (Definitely x)) = Perhaps x
join (Definitely Nope) = Nope
join (Definitely (Perhaps x)) = Perhaps x
join (Definitely (Definitely x)) = Definitely x
Of course, >>= can be defined in terms of join and fmap (which is trivial for Certainty, and can be automatically found by ghc with -XDeriveFunctor) by
x >>= f = join (fmap f x)
which makes my definition equivalent to Carsten's
Moreover, by looking at the last three lines of the definition of join we see that for all x :: Certainty a we have join (Definitely x) = x which suggests that
return x = Definitely x
