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

13 Jul 2017

      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?

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

Francesco Ariis