How works this `do` example?

Previous Topic Next Topic
 
classic Classic list List threaded Threaded
3 messages Options
Baa
Reply | Threaded
Open this post in threaded view
|  
Report Content as Inappropriate

How works this `do` example?

Baa
Hello, Dear List!

Consider, I have:

  request1 :: A -> Connection -> IO ()
  request2 :: A -> Connection -> IO A

How does it work -

  resp <- getConnection
    >>= do request1 myA
           request2 anotherA

?!

It is compiled but seems that does not execute `request1`...

`request1 myA` gets `Connection` value, good. But it does not return
`IO Connection`! It returns `IO ()`. But how does `request2 anotherA`
get `Connection` value too? Because this is not compiled sure:

  resp <- getConnection
    >>= request1 myA >>= request2 anotherA

I tried this:

    module Main where

    f1 :: Int -> IO ()
    f1 i = do
      print "f1!"
      print i
      return ()

    f2 :: Int -> IO Int
    f2 i = do
      print "f2!"
      print i
      return i

    f0 :: IO Int
    f0 = pure 10

    main :: IO ()
    main = f0
      >>= do f1
             f2
      >> print "end"

and I get output:

    "f2!"
    10
    "end"

which means that `f1` is not executing in `do..`-block, but how does `f2` get 10 as input?!


==
Cheers,
  Paul
_______________________________________________
Beginners mailing list
[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Reply | Threaded
Open this post in threaded view
|  
Report Content as Inappropriate

Re: 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?
_______________________________________________
Beginners mailing list
[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Baa
Reply | Threaded
Open this post in threaded view
|  
Report Content as Inappropriate

Re: How works this `do` example?

Baa
I suspected that it was in Read monad, but I don't see where is it this
"Read" monad here :)  Francesco, thank you very much!!

Absolutely clear :)


В Thu, 13 Jul 2017 11:06:13 +0200
Francesco Ariis <[hidden email]> wrote:

> 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?
> _______________________________________________
> Beginners mailing list
> [hidden email]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________
Beginners mailing list
[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Loading...