Haskell Monad - How does Monad on list work?

Question

In order to understand Monad, I came up with the following definitions:

class Applicative\' f where
 purea :: a -> f a
 app :: f (a->b) -> f a ->         


        

Answer 1

Monads are often easier understood with the “mathematical definition”, than with the methods of the Haskell standard class. Namely,

class Applicative' m => Monadd m where
  join :: m (m a) -> m a

Note that you can implement the standard version in terms of this, vice versa:

join mma = mma >>= id

ma >>= f = join (fmap f ma)

For lists, join (aka concat) is particularly simple:

join :: [[a]] -> [a]
join xss = [x | xs <- xss, x <- xs]  -- xss::[[a]], xs::[a]
-- join [[1],[2]] ≡ [1,2]

For the example you find confusing, you'd have

[1,2,3,4] >>= \x->[(x+1)]
  ≡   join $ fmap (\x->[(x+1)]) [1,2,3,4]
  ≡   join [[1+1], [2+1], [3+1], [4+1]]
  ≡   join [[2],[3],[4],[5]]
  ≡   [2,3,4,5]