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Today's Topics:
1. Re: Type signature question (Gesh hseG)
2. Re: Type signature question (Louis-Guillaume Gagnon)
3. Re: Type signature question (Chadda? Fouch?)
4. Re: Why does this list comprehension return an empty list?
(Chadda? Fouch?)
5. Pipe and typeclasses (Emmanuel Surleau)
6. Re: Pipe and typeclasses (Karol Samborski)
7. Re: Shorten this code (Kim-Ee Yeoh)
8. Re: Why does this list comprehension return an empty list?
(Kim-Ee Yeoh)
----------------------------------------------------------------------
Message: 1
Date: Tue, 23 Jul 2013 18:05:51 +0300
From: Gesh hseG <g...@gesh.uni.cx>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Type signature question
Message-ID:
<CACS5XqMp12OvW3vHrVb=XFfH=gt95ytg9s2ngzsrqotji0d...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
On Tue, Jul 23, 2013 at 5:48 PM, Denis Kasak <denis.ka...@gmail.com> wrote:
> On 23 July 2013 16:33, Louis-Guillaume Gagnon
> <louis.guillaume.gag...@gmail.com> wrote:
>>
>> and I don't undestand why the "Eq a =>" shows up.
>
The inference works somewhat like this:
GHC sees the definitions of firstHalf, secondHalf and infers they have
the same type as xs,
and that all three of them must be some lists for them to be passed to
and be returned by take and drop.
GHC sees firstHalf == secondHalf. Since (==) has type Eq a => a -> a
-> Bool, firstHalf (and secondHalf and xs)
must have types which are instances of the typeclass Eq.
Looking at the instance declarations for Eq, we find that for all
types t which are instances of Eq, the type [t]
(lists of values of type t) is also an instance of Eq.
In other words, instance Eq a => Eq [a] where ...
Thus, we have that firstHalf, secondHalf, xs :: Eq t => [t] and cannot
infer a more specific type for these three names,
leaving us with the signature Eq t => [t] -> Bool for isPalindrome.
Intuitively, since you're checking whether the first half of the list
is equal to the second half, you need to be able to
check that the first element is equal to the last, second to the
before-last, etc.
Also, notice that you could just have written
isPalindrome xs = xs == reverse xs
or even
isPalindrome = (==) <$> id <*> reverse
but this last way is too obfuscated for most sane people.
HTH,
Gesh
------------------------------
Message: 2
Date: Tue, 23 Jul 2013 11:21:16 -0400
From: Louis-Guillaume Gagnon <louis.guillaume.gag...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Type signature question
Message-ID:
<CANCBZJvockekLGVU6RqwikxuWe=oas1f1+sgmd3q0gazcqp...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
2013/7/23 Gesh hseG <g...@gesh.uni.cx>:
> The inference works somewhat like this:
> GHC sees the definitions of firstHalf, secondHalf and infers they have
> the same type as xs,
> and that all three of them must be some lists for them to be passed to
> and be returned by take and drop.
> GHC sees firstHalf == secondHalf. Since (==) has type Eq a => a -> a
> -> Bool, firstHalf (and secondHalf and xs)
> must have types which are instances of the typeclass Eq.
> Looking at the instance declarations for Eq, we find that for all
> types t which are instances of Eq, the type [t]
> (lists of values of type t) is also an instance of Eq.
> In other words, instance Eq a => Eq [a] where ...
>
> Thus, we have that firstHalf, secondHalf, xs :: Eq t => [t] and cannot
> infer a more specific type for these three names,
> leaving us with the signature Eq t => [t] -> Bool for isPalindrome.
>
> Intuitively, since you're checking whether the first half of the list
> is equal to the second half, you need to be able to
> check that the first element is equal to the last, second to the
> before-last, etc.
great post, thanks!
> Also, notice that you could just have written
> isPalindrome xs = xs == reverse xs
Why didn't I think of that!?
glg
------------------------------
Message: 3
Date: Tue, 23 Jul 2013 19:48:14 +0200
From: Chadda? Fouch? <chaddai.fou...@gmail.com>
To: beginners <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Type signature question
Message-ID:
<canfjzrbsu25xufjkxxbutywaywakscrcvdfaoiwlfsbz+4u...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Le 23 juil. 2013 17:23, "Louis-Guillaume Gagnon" <
louis.guillaume.gag...@gmail.com> a ?crit :
>
> > Also, notice that you could just have written
> > isPalindrome xs = xs == reverse xs
>
> Why didn't I think of that
Probably because your internal definition of palindromes is slightly wrong
: note that your function doesn't do the same thing as this one, more
precisely it refuse any palindrome of odd length like [1, 2, 3, 2, 1]. Of
course you may have your own definition that makes even length a
requirement.
--
Jeda?
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Message: 4
Date: Tue, 23 Jul 2013 21:12:51 +0200
From: Chadda? Fouch? <chaddai.fou...@gmail.com>
To: beginners <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Why does this list comprehension
return an empty list?
Message-ID:
<CANfjZRY30CTA+L6fV4W=6jpoxxlejin_apxpmyv0wr_tzas...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Le 23 juil. 2013 10:54, "Costello, Roger L." <coste...@mitre.org> a ?crit :
>
> Hi Folks,
>
> I have a list of singletons:
>
> xs = [("a")]
>
> f is a function that, given an argument x, it returns the argument:
>
> f x = x
>
> g is a function that, given an argument x, it returns the empty list:
>
> g x = []
>
> I have a list comprehension that extracts the singletons from xs using f
and g, and creates a pair from their output:
>
> [(a,b) | a <- f xs, b <- g xs]
>
> I executed this and the result is the empty list:
>
> []
>
> That is odd. Why is the empty list the result?
This is pretty normal since there are no elements in "g xs" so b can takes
no values. You have got some excellent answers on that but I think your
problem is more fundamental :
> f x = x
This function does nothing or more precisely it is the identity, you can
replace "f anything" by "anything" and have exactly the same result.
> g x = []
This function just takes anything and returns an empty list. So
> [(a,b) | a <- f xs, b <- g xs]
Is exactly the same as :
> [(a,b) | a <- xs, b <- [] ]
If xs is ["a"] that becomes
> [(a,b) | a <- ["a"], b <- [] ]
I think those f and g didn't work like that in your mind so could you
explain what you thought they should do (with some examples maybe).
--
Jeda?
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Message: 5
Date: Tue, 23 Jul 2013 21:48:06 +0200
From: Emmanuel Surleau <emmanuel.surl...@gmail.com>
To: Haskell Beginners <beginners@haskell.org>
Subject: [Haskell-beginners] Pipe and typeclasses
Message-ID: <20130723194806.GA9606@emm-laptop>
Content-Type: text/plain; charset=us-ascii
Hello,
I have seen from time to time the pipe symbol in type signatures, like so:
class Monad m => MonadReader r m | m -> r where
How should I interpret that ? Or more to the point, how does the compiler
interpret it?
Thanks,
Emm
------------------------------
Message: 6
Date: Tue, 23 Jul 2013 21:56:20 +0200
From: Karol Samborski <edv.ka...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Pipe and typeclasses
Message-ID:
<cace2dtt3hh8sdmw+6+kixrnfxlyp6xyd2hdujihe7dgkndv...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
2013/7/23 Emmanuel Surleau <emmanuel.surl...@gmail.com>
> Hello,
>
> I have seen from time to time the pipe symbol in type signatures, like so:
>
> class Monad m => MonadReader r m | m -> r where
>
> How should I interpret that ? Or more to the point, how does the compiler
> interpret it?
>
Hi Emmanuel!
The pipe symbol is for a functional dependency:
http://www.haskell.org/haskellwiki/Functional_dependencies
Karol
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Message: 7
Date: Wed, 24 Jul 2013 07:36:28 +0700
From: Kim-Ee Yeoh <k...@atamo.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Shorten this code
Message-ID:
<capy+zdrhv8pofdtuu32_derxetyzgqnnxo_rqwh-fufcmkd...@mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"
On Tue, Jul 23, 2013 at 5:41 PM, Nadav Chernin <nadavcher...@gmail.com>wrote:
> Please try to shorten it:
>
What would you achieve by doing so?
-- Kim-Ee
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Message: 8
Date: Wed, 24 Jul 2013 07:47:24 +0700
From: Kim-Ee Yeoh <k...@atamo.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Why does this list comprehension
return an empty list?
Message-ID:
<capy+zdrowvpcuvk5sva11fq9f9+2pntetqgrgbulqfmcnto...@mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"
On Tue, Jul 23, 2013 at 3:53 PM, Costello, Roger L. <coste...@mitre.org>wrote:
> Also, how would I modify this:
>
> [(a,b) | a <- f xs, b <- g xs]
>
> so that it produces this:
>
> [("a",[])]
>
Try: [(a,b) | a <- f xs | b <- g xs]
Note the 2nd parallel bar. You may have to enable a Parallel List
Comprehension pragma or somesuch.
Everyone else has given great explanations about your curious definitions
of f and g and why the original does what it does. Well worth looking into.
-- Kim-Ee
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