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Today's Topics:
1. How works this `do` example? (Baa)
2. Re: How works this `do` example? (Francesco Ariis)
3. Re: How works this `do` example? (Baa)
----------------------------------------------------------------------
Message: 1
Date: Thu, 13 Jul 2017 11:29:56 +0300
From: Baa <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] How works this `do` example?
Message-ID: <20170713112956.15426e6c@Pavel>
Content-Type: text/plain; charset=US-ASCII
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
------------------------------
Message: 2
Date: Thu, 13 Jul 2017 11:06:13 +0200
From: Francesco Ariis <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] How works this `do` example?
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8
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?
------------------------------
Message: 3
Date: Thu, 13 Jul 2017 12:41:49 +0300
From: Baa <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] How works this `do` example?
Message-ID: <20170713124149.069163e2@Pavel>
Content-Type: text/plain; charset=UTF-8
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 <[email protected]> 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
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
------------------------------
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