[Haskell-beginners] How works this `do` example?

[Francesco Ariis]

On Thu, Jul 13, 2017 at 11:29:56AM +0300, Baa wrote:
>     main :: IO ()
>     main = f0
>       >>= do f1
>              f2
>       >> print "end"
> 
> and I get output:
> 
>     "f2!"
>     10
>     "end"

Hello Paul, your `main` desugars to

    main = f0 >>= (f1 >> f2) >> print "end"

Now, the quizzical part is

    λ> :t (f1 >> f2)
    (f1 >> f2) :: Int -> IO Int

Why does this even type checks? Because:

    λ> :i (->)
    [..]
    instance Monad ((->) r) -- Defined in ‘GHC.Base’
    [..]

((->) r) is an instance of Monad! The instance is:

    instance Monad ((->) r) where
        f >>= k = \r -> k (f r) r

you already know that `m >> k` is defined as `m >>= \_ -> k`, so

        f >> k = \r -> (\_ -> k) (f r) r
               = \r -> k r

Is it clear enough?