Re: [Haskell-cafe] Are all monads functions?

In that sense every value in maths is a function. In other words: Your extension of everything (!) to functions is redundant.

And function is not unique in this way. All types can be embedded into pairs also, e.g., newtype MyInt = MyInt ((),Int), or newtype MyInt = MyInt (((),Int),()), etc. - Conal 2011/12/31 Ertugrul Söylemez

Yves Parès 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.

Greets, Ertugrul