Re: [Haskell-cafe] Does Backwards admit a Monad instance?

15 Dec 2015

      I realized this does not quite satisfy the law

pure x >>= f = f x

In particular,

pure () >>= (\() -> m) :: Backwards m

is a divergent computation whenever m is a "strict" monad. Things seem
somewhat nicer when m is "lazy", but there could be further problems
I've overlooked.

On Tue, Dec 15, 2015 at 2:20 PM, David Feuer  wrote:
...
In particular, I'm thinking about
instance MonadFix m => Monad (Backwards m) where
  m >>= f = Backwards $
    do
      fin <- forwards (f int)
      int <- forwards m
      return fin
It looks to me like this should be valid, and compatible with the
Applicative instance, but maybe I'm missing something.