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Today's Topics:
1. Re: Linebreaks in lhs2tex eval
(Sumit Sahrawat, Maths & Computing, IIT (BHU))
2. Type declaration (Christian Sperandio)
3. Re: Type declaration (Brandon Allbery)
4. Operator orders affect parse result (m00nlight)
5. Re: Operator orders affect parse result (Francesco Ariis)
----------------------------------------------------------------------
Message: 1
Date: Sun, 12 Apr 2015 21:09:33 +0530
From: "Sumit Sahrawat, Maths & Computing, IIT (BHU)"
<[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Linebreaks in lhs2tex eval
Message-ID:
<CAJbEW8OumVHthHZ+tUr2p7cy=kv02hx3ak9wv3e_xw44fey...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Try opening an issue here: https://github.com/kosmikus/lhs2tex/issues.
On 12 April 2015 at 17:25, martin <[email protected]> wrote:
> Hello all,
>
> it appears lhs2tex's \eval command does not show linebreaks. This is a
> pity, particularly when the output is formatted
> with a pretty printer.
>
> Is there a way to teach \eval to respect the linebreaks?
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
--
Regards
Sumit Sahrawat
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Message: 2
Date: Sun, 12 Apr 2015 18:41:50 +0200
From: Christian Sperandio <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] Type declaration
Message-ID: <[email protected]>
Content-Type: text/plain; charset="utf-8"
Hi,
I?m currently playing with the mutable array to implement an heap sort sort
(from The Art of Computer Programming) and I?m puzzled with an error.
When I code:
toHeap :: Ord a => [a] -> IO [a]
toHeap [] = return []
toHeap [x] = return [x]
toHeap xs = do
arr <- newListArray (1, length xs) xs :: IO (IOArray Int a)
getElems arr
I?ve got this error:
Could not deduce (a ~ a1)
from the context (Ord a)
bound by the type signature for toHeap :: Ord a => [a] -> IO [a]
at
/var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:9:11-32
?a? is a rigid type variable bound by
the type signature for toHeap :: Ord a => [a] -> IO [a]
at
/var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:9:11
?a1? is a rigid type variable bound by
an expression type signature: IO (IOArray Int a1)
at
/var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:14:10
Expected type: [a1]
Actual type: [a]
Relevant bindings include
xs :: [a]
(bound at
/var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:12:8)
toHeap :: [a] -> IO [a]
(bound at
/var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:10:1)
In the second argument of ?newListArray?, namely ?xs?
In a stmt of a 'do' block:
arr <- newListArray (1, length xs) xs :: IO (IOArray Int a)
But with the code below, all work:
toHeap :: Ord a => [a] -> IO [a]
toHeap [] = return []
toHeap [x] = return [x]
toHeap xs = do
arr <- buildArray xs
getElems arr
buildArray :: [a] -> IO (IOArray Int a)
buildArray xs = newListArray (1, length xs) xs
What do I miss ?
Thanks.
Chris
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Message: 3
Date: Sun, 12 Apr 2015 12:55:05 -0400
From: Brandon Allbery <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Type declaration
Message-ID:
<CAKFCL4V+pFSg_Pi-zyqqwKL6oGJ2MGULx6a8LxMz4j=+t6b...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On Sun, Apr 12, 2015 at 12:41 PM, Christian Sperandio <
[email protected]> wrote:
> I?m currently playing with the mutable array to implement an heap sort
> sort (from The Art of Computer Programming) and I?m puzzled with an error.
>
> When I code:
>
> toHeap :: Ord a => [a] -> IO [a]
> toHeap [] = return []
> toHeap [x] = return [x]
> toHeap xs = do
> arr <- newListArray (1, length xs) xs :: IO (IOArray Int a)
> getElems arr
>
Note that the "a" in the signature "IO (IOArray Int a)" is *not* the same
as the one in the signature of toHeap; the scope of that type variable is
the signature itself, not the following equation(s). You have in effect
done the opposite of what you intended --- instead of asserting it is the
same, you asserted that it is a *different* one, by handing the compiler an
unexpected `a` which must be assumed to represent a distinct type.
If you need to extend the scope of a type variable like this, you need the
ScopedTypeVariables extension, and to declare the type variable as having
extended scope with an explicit `forall`:
{-# LANGUAGE ScopedTypeVariables #-}
toHeap :: forall a. Ord a => [a] -> IO [a]
toHeap [] = return []
toHeap [x] = return [x]
toHeap xs = do
arr <- newListArray (1, length xs) xs :: IO (IOArray Int a)
getElems arr
--
brandon s allbery kf8nh sine nomine associates
[email protected] [email protected]
unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
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Message: 4
Date: Mon, 13 Apr 2015 14:56:47 +0800 (CST)
From: m00nlight <[email protected]>
To: "The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell" <[email protected]>
Subject: [Haskell-beginners] Operator orders affect parse result
Message-ID: <[email protected]>
Content-Type: text/plain; charset="utf-8"
Dear Haskellers,
I have tried the "while language parser" tutorial on haskellwiki, but I
encounter an problem I can not understand here,
The following is the code:
http://pastebin.com/saC9vwCn
The problem is, with the code, the parse result of "fact := a / b * c" is
*Main> parseString "fact := a / b * c"
Assign "fact" (ABinary Multiply (ABinary Divide (Var "a") (Var "b")) (Var "c"))
*Main>
But if I swap the order of "/" and "*" operators in aOperators(make "*" appear
before "/" operator), then the result is
*Main> parseString "fact := a / b * c"
Assign "fact" (ABinary Divide (Var "a") (ABinary Multiply (Var "b") (Var "c")))
*Main>
So it seems that the order of the operator will affect the parse result. What's
the problem with my code? How can I
make my program to produce consistent parse result.
Best Regards,
--m00nlight
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Message: 5
Date: Mon, 13 Apr 2015 09:21:19 +0200
From: Francesco Ariis <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] Operator orders affect parse result
Message-ID: <[email protected]>
Content-Type: text/plain; charset=us-ascii
On Mon, Apr 13, 2015 at 02:56:47PM +0800, m00nlight wrote:
> Dear Haskellers,
>
> I have tried the "while language parser" tutorial on haskellwiki, but I
> encounter an problem I can not understand here,
> The following is the code:
>
> http://pastebin.com/saC9vwCn
>
>
> So it seems that the order of the operator will affect the parse result.
> What's the problem with my code? How can I make my program to produce
> consistent parse result.
Checking the documentation for OperatorTable [1] (the first argument to
buildExpressionParser), it seems it is working as expected:
An OperatorTable s u m a is a list of Operator s u m a lists. The list
is ordered in descending precedence. All operators in one list have
the same precedence (but may have a different associativity).
I think you can take advantage of the fact that OperatorTable is /a list
of lists/, hence
aOperators = [ [Prefix (reservedOp "-" >> return (Neg ))]
, [Infix (reservedOp "/" >> return (ABinary Divide ))
AssocLeft,
Infix (reservedOp "*" >> return (ABinary Multiply ))
AssocLeft]
, [Infix (reservedOp "+" >> return (ABinary Add )) AssocLeft]
, [Infix (reservedOp "-" >> return (ABinary Subtract ))
AssocLeft]
]
should do the trick (try swapping the elements in the second list.
Will this do?
[1]
http://hackage.haskell.org/package/parsec-3.0.0/docs/Text-Parsec-Expr.html#t:OperatorTable
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