[Haskell-cafe] Are all monads functions?
es at ertes.de
Sun Jan 1 05:17:12 CET 2012
Yves Parès <limestrael+haskell at gmail.com> wrote:
> But still, I maintain my previous view. I could clarify that by saying
> that (e.g. for Maybe) we could separate it in two types, Maybe itself
> and its monad:
> -- The plain Maybe type
> data Maybe a = Just a | Nothing
> -- The MaybeMonad
> newtype MaybeMonad a = MM ( () -> Maybe a )
> That's what using Maybe as a monad semantically means, doesn't it?
That's a statement like "the sky is blue". You can represent any value
as a function of (). You are saying that every integer is a function.
newtype MyInt = MyInt (() -> Int)
newtype My a = My (() -> a)
Think of it this way: There is something like a canonical
representation of every monad. If you let that one be the one with the
least order (which is reasonable), then no, not every monad's canonical
representation is a function, because the base library definition of
Maybe is the canonical one (order zero).
In that sense every value in maths is a function. In other words: Your
extension of everything (!) to functions is redundant.
You get the idea.
nightmare = unsafePerformIO (getWrongWife >>= sex)
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