Using return vs. not using return in the list monad

Question

I started my Grand Haskell Crusade (GHC :) ) and I am a bit confused with monads and IO functions. Could anyone explain simply what is the difference between those

Answer 1

Desugaring the do syntax to the equivalent

f1 = [1,2] >>= \x -> [x, x+1]
f2 = [1,2] >>= \x -> return [x, x+1]

Now, >>= comes from the Monad class,

class Monad m where
    (>>=) :: m a -> (a -> m b) -> m b
    return :: a -> m a

and the LHS of >>= in both f1 and f2 is [a] (where a has defaulted to Integer), so we're really considering

instance Monad [] where
    (>>=) :: [a] -> (a -> [b]) -> [b]
    ...

This obeys the same monad laws as, but is a different monad from,

instance Monad IO where ...

and >>= for other monads, so don't just blindly apply what you know about one to the other, okay? :)

instance Monad [] is defined thusly in GHC

instance Monad [] where
    m >>= k = foldr ((++) . k) [] m
    return x = [x]
    ...

but it is probably easier to understand []'s >>= as

instance Monad [] where
    m >>= k = concatMap k m

If you take this and apply it to the original, you get

f1 = concatMap (\x -> [x, x+1]) [1,2]
f2 = concatMap (\x -> [[x, x+1]]) [1,2]

and it's clear why the values of f1 and f2 are what they are.