I understand the definition of >>= in term of join
xs >>= f = join (fmap f xs)
which also tells us that fmap + join yields >>=
I was wondering if for the List monad it's possible to define without join, as we do for example for Maybe:
>>= m f = case m of
Nothing -> Nothing
Just x -> f x
Sure. The actual definition in GHC/Base.hs is in terms of the equivalent list comprehension:
instance Monad [] where
xs >>= f = [y | x <- xs, y <- f x]
Alternatively, you could try the following method of working it out from scratch from the type:
(>>=) :: [a] -> (a -> [b]) -> [b]
We need to handle two cases:
[] >>= f = ???
(x:xs) >>= f = ???
The first is easy. We have no elements of type a, so we can't apply f. The only thing we can do is return an empty list:
[] >>= f = []
For the second, x is a value of type a, so we can apply f giving us a value of f x of type [b]. That's the beginning of our list, and we can concatenate it with the rest of the list generated by a recursive call:
(x:xs) >>= f = f x ++ (xs >>= f)
