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Today's Topics:
1. Re: Beginners Digest, Vol 56, Issue 33 (Brandon Allbery)
2. Re: Beginners Digest, Vol 56, Issue 33 (Brent Yorgey)
3. Re: Use a list as data for a data type? (Karl Voelker)
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Message: 1
Date: Fri, 22 Feb 2013 19:42:06 -0500
From: Brandon Allbery <[email protected]>
Subject: Re: [Haskell-beginners] Beginners Digest, Vol 56, Issue 33
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Message-ID:
<CAKFCL4XKfsLQLKro61ey=_Kxa=j9taz7k_yo13s8akhvlpo...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On Fri, Feb 22, 2013 at 7:27 PM, xiao Ling <[email protected]> wrote:
> h :: M Int -> M Int -> M Int
> h x y = bind ( \x-> g x y ) x
>
> where g is
>
> g :: Int -> W Int -> W Int
> g x y = y >>= (return . (+x))
>
> for the monad:
>
> data M a = M a deriving Show
>
> Now I am a little confused, how can you put x in g if it takes an Int as
> first parameter but x is M Int?
>
Because it's a different "x". Lemme rewrite it slightly:
h :: M Int -> M Int -> M Int
h x y = bind ( \w -> g w y ) x
All I did was replace the inner "x" with "w", to demonstrate that it has no
relationship to the outer "x"; the \... -> syntax introduces new local
bindings unrelated to any outside of it, in this case for "w" (or what he
had "x", shadowing the original binding of "x" within the lambda).
--
brandon s allbery kf8nh sine nomine associates
[email protected] [email protected]
unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
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Message: 2
Date: Fri, 22 Feb 2013 19:44:05 -0500
From: Brent Yorgey <[email protected]>
Subject: Re: [Haskell-beginners] Beginners Digest, Vol 56, Issue 33
To: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset=us-ascii
On Fri, Feb 22, 2013 at 07:27:28PM -0500, xiao Ling wrote:
> Hi All: How do you define a function of signature h :: M Int -> M Int -> M
> Int so that h ( M x ) ( M y ) = M ( x + y ), but without unwrapping the
> value from the monad?
>
> This question is from the article "Trivial Monad" found at
> http://blog.sigfpe.com/2007/04/trivial-monad.html. The provided answer is
>
> h x y = x >>= (\x -> g x y)
>
> Now I am a little confused, how can you put x in g if it takes an Int as
> first parameter but x is M Int?
I think your confusion may stem from the fact that there are *two
different* things named 'x' in the above code. The x's on the right
hand side of the >>= shadow the x on the left hand side. This is
confusing and poor style; a better way to write it would be
h x y = x >>= (\i -> g i y)
If you study the type of >>= you will see that i indeed has type Int,
as required for the first argument of g.
-Brent
------------------------------
Message: 3
Date: Fri, 22 Feb 2013 19:48:20 -0800
From: Karl Voelker <[email protected]>
Subject: Re: [Haskell-beginners] Use a list as data for a data type?
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Message-ID:
<CAFfow0wzUMVne5=j1675ul0uujyw4a3vaaukahp9c86sjxa...@mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"
On Fri, Feb 22, 2013 at 12:40 PM, Bryce Verdier <[email protected]>wrote:
> Brent,
>
> Thank you for the responses. I appreciate it. Bummer about there not being
> some kind of built-in/generic shortcut.
>
> Bryce
I am curious about what motivated your question, since it's hard to imagine
a simpler alternative to writing "IpAddress 1 2 3 4".
-Karl Voelker
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