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Today's Topics:
1. pattern matching on date type (Max.cs)
2. Re: pattern matching on date type (Alexander Dunlap)
3. Re: about the concatenation on a tree (Thomas Davie)
4. Re: pattern matching on date type (Thomas Davie)
5. Re: until and Time (Steve Klabnik)
6. Question about typeclasses (keke)
7. Re: Question about typeclasses (Jason Dusek)
----------------------------------------------------------------------
Message: 1
Date: Thu, 1 Jan 2009 08:36:24 -0000
From: "Max.cs" <[email protected]>
Subject: [Haskell-beginners] pattern matching on date type
To: "Brandon S. Allbery KF8NH" <[email protected]>,
<[email protected]>, <[email protected]>
Message-ID: <3d524ba96876446ca9b57bdb8e096...@hxzhao>
Content-Type: text/plain; charset="iso-8859-1"
thanks!
suppose we have
> data Tree a = Leaf a | Branch (Tree a) (Tree a) deriving Show
and how I could define a function foo :: a -> Tree a that
foo a = Leaf a where a is not a type of Tree
foo b = b where b is one of the type of Tree (Leaf or Branch) ?
The following code seems not working......
foo (Leaf a) = a
foo a = Leaf a
saying 'Couldn't match expected type `a' against inferred type `Btree a''
any idea?
Thanks,
Max
From: Brandon S. Allbery KF8NH
Sent: Thursday, January 01, 2009 7:35 AM
To: Max.cs
Cc: Brandon S. Allbery KF8NH ; [email protected] ; [email protected]
Subject: Re: [Haskell-cafe] definition of data
On 2009 Jan 1, at 2:32, Max.cs wrote:
data Tree a = a | Branch (Tree a) (Tree a) deriving Show
but it seems not accpetable in haskell ?
You need a constructor in both legs of the type:
> data Tree a = Leaf a | Branch (Tree a) (Tree a) deriving Show
--
brandon s. allbery [solaris,freebsd,perl,pugs,haskell] [email protected]
system administrator [openafs,heimdal,too many hats] [email protected]
electrical and computer engineering, carnegie mellon university KF8NH
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Message: 2
Date: Thu, 1 Jan 2009 01:44:33 -0800
From: "Alexander Dunlap" <[email protected]>
Subject: Re: [Haskell-beginners] pattern matching on date type
To: Max.cs <[email protected]>
Cc: [email protected], [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset=ISO-8859-1
On Thu, Jan 1, 2009 at 12:36 AM, Max.cs <[email protected]> wrote:
> thanks!
>
> suppose we have
>
>> data Tree a = Leaf a | Branch (Tree a) (Tree a) deriving Show
>
> and how I could define a function foo :: a -> Tree a that
>
> foo a = Leaf a where a is not a type of Tree
> foo b = b where b is one of the type of Tree (Leaf or
> Branch) ?
>
> The following code seems not working......
>
> foo (Leaf a) = a
> foo a = Leaf a
>
> saying 'Couldn't match expected type `a' against inferred type `Btree a''
>
> any idea?
>
> Thanks,
> Max
You can't define such a function. foo :: a -> Tree a, but the
definition foo b = b has the type a -> a, which is why the compiler
says it can't match type "a" against "Tree a". In general, Haskell
functions can't "look" at the type of their arguments: they are either
monomorphic or are polymorphic, in which case you can only use
polymorphic functions that match the polymorphic type. For the problem
you are trying to solve, you probably need to encode this logic higher
up in the overall function (e.g. have one function to deal with "a"s
and another to deal with "Tree a"s).
Hope that helps,
Alex
------------------------------
Message: 3
Date: Thu, 1 Jan 2009 10:58:26 +0100
From: Thomas Davie <[email protected]>
Subject: [Haskell-beginners] Re: about the concatenation on a tree
To: "Max.cs" <[email protected]>
Cc: Beginners Haskell <[email protected]>
Message-ID: <[email protected]>
Content-Type: text/plain; charset=WINDOWS-1252; format=flowed;
delsp=yes
>
> hi Bob,
>
> I believe the type of foldTree is implementable, but what confuses
> me is
>
> the output example:
>
>> concatT (Leaf t) = t
>> concatT (Branch (Leaf t1) (Leaf t2)) = Branch t1 t2
>
> they seem to be inconsistent with the type of concatT which return a
> Tree only, but in the Leaf case, a atomic value is returned?
>
> anything important I am missing?
You need to remember what type is stored in that Leaf remember that
this is a Tree (Tree a) i.e. the variable t in the line above has
the type Tree a.
Bob
------------------------------
Message: 4
Date: Thu, 1 Jan 2009 11:04:23 +0100
From: Thomas Davie <[email protected]>
Subject: Re: [Haskell-beginners] pattern matching on date type
To: "Max.cs" <[email protected]>
Cc: [email protected], [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset=WINDOWS-1252; format=flowed;
delsp=yes
On 1 Jan 2009, at 09:36, Max.cs wrote:
> thanks!
>
> suppose we have
>
> > data Tree a = Leaf a | Branch (Tree a) (Tree a) deriving Show
>
> and how I could define a function foo :: a -> Tree a that
>
> foo a = Leaf a where a is not a type of Tree
> foo b = b where b is one of the type of Tree (Leaf
> or Branch) ?
>
> The following code seems not working......
>
> foo (Leaf a) = a
> foo a = Leaf a
>
> saying 'Couldn't match expected type `a' against inferred type
> `Btree a'
Hi again Max,
I'm assuming this is continuing from the concatT example, and that
you're struggling with first function you must pass to foldTree.
Remember the type of the function it's not a -> Tree a, but Tree a -
> Tree a, because your leaves in the parent tree all contain trees to
glue on at that point.
So, the function you want, is the function which looks at the
parameter it's given, goes "oh, that's interesting", does nothing to
it, and hands it back to replace the Leaf. I recommend searching
hoogle for functions of type a -> a, the function you're looking for
is built in.
Bob
------------------------------
Message: 5
Date: Thu, 1 Jan 2009 12:28:00 -0800
From: "Steve Klabnik" <[email protected]>
Subject: Re: [Haskell-beginners] until and Time
To: "Daniel Fischer" <[email protected]>
Cc: [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="iso-8859-1"
Ahh. Thank you. That makes much more sense now.
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Message: 6
Date: Sat, 3 Jan 2009 16:17:17 +0800
From: keke <[email protected]>
Subject: [Haskell-beginners] Question about typeclasses
To: [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset="utf-8"
Hi,
I am bit struggling with understanding the typeclasses usage in
Text.Regex.Posix (=~) while reading Real world haskell.
The type of (=~) is (RegexMaker Regex CompOption ExecOption source,
RegexContext Regex source1 target) => source1 -> source -> target
I am from Java world. My question is that can I understand source1 in the
way which is a value has type of RegexMaker, Regex, CompOption and
ExecOption?
And the definiton of instance RegexMaker Regex CompOption ExecOption
Stringmakes it possible for us to pass a String as the parameter of
=~?
Where can I find some good metarials about GHC's type classes? I googled a
lot but can not find something mentioned above usage.
--
Cheers,
Keke
-----------------
We paranoid love life
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Message: 7
Date: Sat, 3 Jan 2009 01:26:23 -0800
From: "Jason Dusek" <[email protected]>
Subject: Re: [Haskell-beginners] Question about typeclasses
To: keke <[email protected]>
Cc: [email protected]
Message-ID:
<[email protected]>
Content-Type: text/plain; charset=UTF-8
<[email protected]> 2009-01-03T08:17:00Z
> I am bit struggling with understanding the typeclasses usage
> in Text.Regex.Posix (=~) [...] My question is that can I
> understand source1 in the way which is a value has type of
> RegexMaker, Regex, CompOption and ExecOption?
Prettifying the signature a little bit:
:: ( RegexMaker Regex CompOption ExecOption source
, RegexContext Regex source1 target )
=> source1
-> source
-> target
You can read this as:
The type `source1 -> source -> target` such that both "class
constraints" are satisfied.
What is a class contraint and what does it mean for it to be
satisfied? A class contraint is something like an interface,
or a generic, or a template. One defines "instances" of
classes in Haskell, as in Jave one "implements" an interface.
So what do these particular classes mean? The first one reads:
We have a way to make a regular expression using the usual
option specifications (multiline, case insensitive, stuff
like that -- those are `CompOption` and `ExecOption`).
The second one reads
There is a way to match the first value against the second
to get the desired target value (list of matches, Boolean,
&c.).
The "way to..." in each case is a function in the class
definition -- `makeRegex` in the former case, and `match` in
the latter.
--
Jason Dusek
------------------------------
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