Re: [Haskell-cafe] Newbie: Applying Unknown Number Arguments to A Partial Function
Chris, the subject states clearly that Aditya is a Newbie, and is most likely just trying to define the function map. So I think pointing to a bunch of advanced type magic tricks is not really helpful. Aditya, you say you want the function applyArgument to take a function and a list and apply the function to all elements of that list. The result of applyArgument is a list, with the results of applying the function to each element. So you want the type of applyArgument to be: applyArgument :: (a-b) - [a] - [b] There are a numbers of errors in your code. applyArgument f (arg) = f arg The variable f is the first parameter of applyArgument, and has type (a - b) The variable arg is the second parameter of applyArgument, and has type [a] . You try to apply f to a list which is not ok. Most likely you meant [arg], which is a singleton list, instead of (arg), which is the same as just arg Furthermore applyArgument returns a list of result. The function f only yields a b, instead of a list [b] The following does work: applyArgument f [arg] = [f arg] In your second line: applyArgument f (arg:args) = applyArgument (f arg) args you use a list pattern (arg:args), which is good. The variable arg is the head of the list, and the variable args is the tail of the list. So arg has type a, and args has type [a] . The application (f arg) has type b. Because the function applyArgument expects a function as first argument and not a b, so this is wrong. I guess you can find out by your self how to fix this. There are a number of very good tutorials for beginners on http://haskell.org . Finally your function won't work for empty lists, it is only defined for singleton lists and longer lists. You can fix this by replacing the definition for singleton lists with a definition for the empty list. The pattern for empty list is simply []. Good luck, Arthur On 19-mei-06, at 10:10, Chris Kuklewicz wrote: Aditya Siram wrote: I am trying to write a function 'applyArguments' which takes a function and a list and recursively uses element each in the list as an argument to the function. I want to do this for any function taking any number of arguments. applyArgument f (arg) = f arg applyArgument f (arg:args) = applyArgument (f arg) args This has failed in Hugs, so my question is: Can I conceptually do this? If so, what is the type signature of this function? Deech You can't do that, but there are other tricks that do work: http://okmij.org/ftp/Haskell/types.html which describes Functions with the variable number of (variously typed) arguments and Genuine keyword arguments -- Chris ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Newbie: Applying Unknown Number Arguments to A Partial Function
My apologies to Chris, I think I misinterpreted Aditya's description. Thanks to David House for telling me. I thought he was describing a function such as map instead of polyvaric functions, which would have been more likely for a newbie :-) So to answer Aditya's question, whether you can do this in Haskell. The short answer is no. You cannot do this in plain Haskell. However using various extensions to Haskell, you can indeed use the smart tricks as Chris pointed out: http://okmij.org/ftp/Haskell/types.html Cheers, Arthur On 19-mei-06, at 11:24, Arthur Baars wrote: Chris, the subject states clearly that Aditya is a Newbie, and is most likely just trying to define the function map. So I think pointing to a bunch of advanced type magic tricks is not really helpful. Aditya, you say you want the function applyArgument to take a function and a list and apply the function to all elements of that list. The result of applyArgument is a list, with the results of applying the function to each element. So you want the type of applyArgument to be: applyArgument :: (a-b) - [a] - [b] There are a numbers of errors in your code. applyArgument f (arg) = f arg The variable f is the first parameter of applyArgument, and has type (a - b) The variable arg is the second parameter of applyArgument, and has type [a] . You try to apply f to a list which is not ok. Most likely you meant [arg], which is a singleton list, instead of (arg), which is the same as just arg Furthermore applyArgument returns a list of result. The function f only yields a b, instead of a list [b] The following does work: applyArgument f [arg] = [f arg] In your second line: applyArgument f (arg:args) = applyArgument (f arg) args you use a list pattern (arg:args), which is good. The variable arg is the head of the list, and the variable args is the tail of the list. So arg has type a, and args has type [a] . The application (f arg) has type b. Because the function applyArgument expects a function as first argument and not a b, so this is wrong. I guess you can find out by your self how to fix this. There are a number of very good tutorials for beginners on http://haskell.org . Finally your function won't work for empty lists, it is only defined for singleton lists and longer lists. You can fix this by replacing the definition for singleton lists with a definition for the empty list. The pattern for empty list is simply []. Good luck, Arthur On 19-mei-06, at 10:10, Chris Kuklewicz wrote: Aditya Siram wrote: I am trying to write a function 'applyArguments' which takes a function and a list and recursively uses element each in the list as an argument to the function. I want to do this for any function taking any number of arguments. applyArgument f (arg) = f arg applyArgument f (arg:args) = applyArgument (f arg) args This has failed in Hugs, so my question is: Can I conceptually do this? If so, what is the type signature of this function? Deech You can't do that, but there are other tricks that do work: http://okmij.org/ftp/Haskell/types.html which describes Functions with the variable number of (variously typed) arguments and Genuine keyword arguments -- Chris ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Small syntax question
The problem is that the semicolon after let a = 3 is consumed as part
of the let-declaration.
To make sure that the semicolon is not parsed as part of the let, you
need to indent it less than the variable a.
For example:
module Main where
main = do {
let a = 3
; return ();
};
Arthur
Why isn't this one [valid]?
module Main where
main = do {
let a = 3;
return ();
};
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Re: [Haskell] Compiling wxHaskell from source?
Wouter Swierstra succeeded compiling wxhaskell for Mac OSX Tiger. See his messages on the wxhaskell-mailinglist: http://sourceforge.net/mailarchive/forum.php?forum_id=34197 From: Wouter Swierstra [EMAIL PROTECTED].Nott.AC.UK Re: Installation on Mac OS X 2005-11-09 04:42 Yep, that fixes it. Thanks again for everyone's help. --Wouter For the record, here's what I had to do Download wxMac2.6.2 cd wxMac-2.6.2/ mkdir mybuild cd mybuild ./configure --disable-shared --disable-unicode make sudo make install Download and unpack wxHaskell 0.9.4 comment out the function wxColour_GetPixel in wxhaskell-0.9.4/wxc/src/ ewxw/eljcolour.cpp ./configure --wx-config=/usr/local/bin/wx-config make make install Cheers, Arthur On 17-nov-05, at 22:29, Joel Reymont wrote: I cannot build it on Mac OSX either. I think it has to do with GCC 4.x. On Nov 17, 2005, at 9:09 PM, Deling Ren wrote: Hi there, is there anyone out there having luck with compiling wxHaskell from source on Debian? My Debian version is 3.1, ghc 6.4.1 from sid. wxWidgets 2.6.3. wxHaskell version is 0.9.4. gcc version is 4.0.3. I got the following error messages when trying to run make. ___ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell
Re: type annotations for GADTs
I think it is a bug.
It also works with silly type annotations:
doy :: Y X a b - Y X a b
doy (Y (X :: foo)(X :: bar)) = Y X X
Looks like something smart is happening in the Pattern Annotation rule.
Arthur
On 29-sep-05, at 12:40, Ganesh Sittampalam wrote:
Hi,
I found it slightly surprising that the program below (also attached
as a file) needs the nested type annotations in the place commented.
It's not really clear whether it's a bug or just a natural consequence
of GADT type-checking, but since Simon Marlow also found it surprising
it at first glance, I thought I'd send it in for comment.
The intuitive argument I came up with for why it might be necessary is
that each GADT requires some kind of type signature in argument
position to generalise its type correctly, and the overall signature
for doy isn't enough to generalise the type of the X arguments.
Cheers,
Ganesh
data X a b where
X :: X a a
data Y x a b where
Y :: x a b - x b c - Y x a c
{- This needs the nested type annotations -}
doy :: Y X a b - Y X a b
doy (Y (X :: X a c) (X :: X c b)) = Y X X
data YX a b where
YX :: X a b - X b c - YX a c
{- This version doesn't need them -}
doyx :: YX a b - YX a b
doyx (YX X X) = YX X
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Re: [Haskell-cafe] mistake in Gentle Introduction
On 22-sep-05, at 11:20, Tomasz Zielonka wrote:
On 9/22/05, Simon Marlow [EMAIL PROTECTED] wrote:
The error message on this is really poor: parse error on input
`length'
An addition like expecting '(' might be more helpful.
Not easy to do - some significant work on Happy would be required (or
rewriting GHC's parser, which is even more unlikely).
However we *would* like to do some simple error recovery in the parser
at some point. The main reason for doing this is to get better
results
from checking partially complete modules in Visual Haskell or hIDE.
How about using the parser made by Arthur Baars, trading speed for
better error messages?
Maybe even the speed issue could be solved.
Speed, error messages and recovery were not my main concern when
writing the parser. I think a lot can be gained by optimizing the
combinator-based parsers. However this is no fun. I had to do some
left-factorization to gain reasonable efficiency(at first the speed was
sometimes worse than 1 token per minute :-)). For a complex language
such as Haskell, the code resulting from factorization is really messy
and the notational elegance of the combinators is lost. Maybe someone
can write a parser-generator that uses Doaitses parsing-engine as a
back-end :-)
A main feature of Doaitses parsing-combinators is that they can recover
from an error in the input. This is done by considering a number of
alternatives to repair the input, and taking the best one to
continue parsing. This proces can be improved by tweaking the insert
and delete costs of certain symbols. Furthermore the error messages can
be customized as with Parsec.
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Re: [Haskell-cafe] wxHaskell: getting a checkbox state
Hi, A Checkbox is instance of the class Checkable: http://wxhaskell.sourceforge.net/doc/ Graphics.UI.WX.Classes.html#t%3ACheckable This means you can get and set the checked property for checkboxes. for example: c - get cbEdit checked set cbEdit [checked := not c ] The following code makes the checkbox print its state every time it is checked or unchecked: cbEdit - checkBox f [text := Edit Mode ] set cbEdit [ on command := do v - get cbEdit checked print v ] Arthur PS: there is a mailinglist for wxhaskell-related questions: http://lists.sourceforge.net/lists/listinfo/wxhaskell-users On 14-sep-05, at 17:15, Mark Carter wrote: I'm a complete n00b to Haskell, and I am trying to write an experimental app using wxHaskell. I'm getting on suprisingly well, given that I have practially no idea what I'm doing. In my main loop I have cbEdit - checkBox p1 [text := Edit Mode, on command := onCbEdit textlog ] where p1 is a panel, and textlog is a textCtrl that I want to write output to. In my where clause I have defined: onCbEdit textlog = do appendText textlog onCbEdit\n So far, this works. The thing that's got me totally stumped is this: suppose I want onCbEdit to do something dependent on the state of the checkbox (i.e. whether it is checked or unchecked). I've tried a variety of things, but I can't get anything to work. To put things into a wider context, here's the message I posted to fa.haskell yesterday: I'm the rawest of n00bs to Haskell. I was interested in evaluating it, and I've seen some GUI stuff that looked pretty impressive. I wanted to write a protoype for a commercial application to do hydrocarbon accounting. Which is to say, you've a number of so-called streams - think of them as graph nodes - feeding into each other in a directed fashion (assume the graph is acyclic). Then there's a bunch of business logic which determines how the values of one stream are affected by the values in another. An example of a stream might be a platform. Another one might be a well. Each stream would display on the window as some funky symbol, with its name underneath. It would be nice if the application had an edit mode, where you could add, move, delete, and join streams in a window; and a data mode, where you could click on the stream you were interested in and examine its properties. Is there a Haskell GUI toolkit which is particularly suitable for this? Other requirements of the toolkit would be: * Works on Windows (XP) * commercial-friendly licence: GPL is out, and even LGPL is a bit fiddly * free * bonus points for being easy to use, and well-documented ___ To help you stay safe and secure online, we've developed the all new Yahoo! Security Centre. http://uk.security.yahoo.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
GADT: weird error message
In the code below the function trans is accepted by GHC 6.4, but trans1 is not. I would expect that (x,y) is just syntactic sugar for (,) x y, but apparently it isn't. I guess this is a bug; can anyone explain what is going on? Cheers, Arthur The Code: data Equal a b where Eq :: Equal a a trans :: forall a b c. Equal a b - Equal b c - Equal a c trans = \x - \y - case (,) x y of (Eq,Eq ) - Eq trans1 :: forall a b c. Equal a b - Equal b c - Equal a c trans1 = \x - \y - case (x, y) of (Eq,Eq ) - Eq The error message: Test2.hs:9:0: Quantified type variable `c' is unified with another quantified type variable a When trying to generalise the type inferred for `trans1' Signature type: forall a b c. Equal a b - Equal b c - Equal a c Type to generalise: Equal a b - Equal b a - Equal a a In the type signature for `trans1' When generalising the type(s) for `trans1' ___ Glasgow-haskell-bugs mailing list Glasgow-haskell-bugs@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
GADT: weird error message
In the code below the function trans is accepted by GHC 6.4, but trans1 is not. I would expect that (x,y) is just syntactic sugar for (,) x y, but apparently it isn't. I guess this is a bug; can anyone explain what is going on? Cheers, Arthur The Code: data Equal a b where Eq :: Equal a a trans :: forall a b c. Equal a b - Equal b c - Equal a c trans = \x - \y - case (,) x y of (Eq,Eq ) - Eq trans1 :: forall a b c. Equal a b - Equal b c - Equal a c trans1 = \x - \y - case (x, y) of (Eq,Eq ) - Eq The error message: Test2.hs:9:0: Quantified type variable `c' is unified with another quantified type variable a When trying to generalise the type inferred for `trans1' Signature type: forall a b c. Equal a b - Equal b c - Equal a c Type to generalise: Equal a b - Equal b a - Equal a a In the type signature for `trans1' When generalising the type(s) for `trans1' ___ Glasgow-haskell-bugs mailing list Glasgow-haskell-bugs@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
Re: [Haskell] ANNOUNCE: ghc-src version 0.2.0
You can check them out using CVS as follows: cvs -d:pserver:[EMAIL PROTECTED]:/data/cvs-rep login cvs -d:pserver:[EMAIL PROTECTED]:/data/cvs-rep checkout uust I'll ask Doaitse to add this information to the web page. Arthur On 30-aug-05, at 13:53, Benjamin Franksen wrote: On Tuesday 30 August 2005 13:04, Arthur Baars wrote: Daan is right, I wrote a parser for GHC using Doaitse Swierstra's parsing combinator library (http://www.cs.uu.nl/groups/ST/Software/UU_Parsing/index.html). Very interesting. I tried to download it but I had no success. How exactly do I checkout the whole uust tree? Ben ___ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell ___ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell
Re: [Haskell-cafe] Parse error
You need { } around the declaration of matrix1. Otherwise the semicolon
at the end of its definition is considered to be part of the 'let':
let { matrix1 = (array ((1,1),(3,3)) [((1,1), 0.0), ((1,2), 0.0),
((1,3),-2.0), ((2,1), 0.0), ((2,2), 7.0), ((2,3), 0.0), ((3,1), 0),
((3,2), 0), ((3,3), -3)])} ;
The layout rule will otherwise add { } as follows:
do { let x = 10 ;
putStr hello ;
}
--
do { let { x = 10 ; empty decl }
putStr hello ;
}
which is wrong, because there is no semicolon separating the
'let'-declaration from the 'putStr' expression.
A solution is to use layout instead of { ; } or write semicolons in
front of declarations:
do { let x = 10
; putStr hello
}
This translates into:
do { let { x = 10 }
; putStr hello
}
The semicolon is not part of the 'let' declaration, because it is
indented less than 'x' .
For more information, see:
http://haskell.org/onlinereport/syntax-iso.html#layout
Hope this helps,
Arthur
On 17-mrt-05, at 20:42, Dmitri Pissarenko wrote:
Hello!
In the attachment you will find a file, in which I try to access Java
from
Haskell using the Java bridge for functional languages.
For details see
http://sourceforge.net/projects/jvm-bridge/
and
http://dapissarenko.com/resources/2005_02_17_eigenvaluesJava/
2005_02_17_eigenva
luesJava.pdf
When I try to compile the attached file using
ghc +RTS -c -RTS -package javavm -c EigenvalueCalculatorHaskell.hs -o
EigenvalueCalculatorHaskell.o
I get the error
EigenvalueCalculatorHaskell.hs:28: parse error on input `putStrLn'
Unfortunately, I have not the slightest idea about how to fix/isolate
it (I
already commented out almost the whole code).
Please tell me what I could try to correct the error. I appreciate ANY
hint.
Many thanks in advance
Dmitri Pissarenko
PS: The source code of the files related to
EigenvalueCalculatorHaskell.hs is
located at
http://dapissarenko.com/resources/2005_02_17_eigenvaluesJava/
2005_02_17_lik.zip
--
Dmitri Pissarenko
Software Engineer
http://dapissarenko.com
EigenvalueCalculatorHaskell.hs___
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Re: [Haskell-cafe] Parser problem continued
The layout of your code is very important when writing haskell code:
Your code :
expr = do t - term
do symbol +
e - expr
return e
return (t + e)
+++ return t
is equivalent to:
expr = do { t - term
; do { symbol +
; e - expr
; return e
}
; return (t + e) -- e is not in scope
}
+++ return t -- t is not in scope
Both t and e are not in scope:
* e is in a nested do-block
* the expression 'return t' is outside the do-block
What you probably mean is:
expr = do t - term
do symbol +
e - expr
return (t + e)
+++ return t
which is equivalent to:
expr = do { t - term
; do { symbol +
; e - expr -- (dropped the return e line)
; return (t + e)
}
+++ return t
}
Now t and e are in scope. The parser 'expr' will first recognize a
'term' and then try to recognize a '+' symbol followed by an
expression. If that fails it returns 't'.
Cheers,
Arthur
On 15-mrt-05, at 16:28, Nicola Whitehead wrote:
Curiouser and curiouser...
expr :: Parser Int
expr = do t - term
do symbol +
e - expr
return (t + e)
+++ return t
solves the undefined variable problem but introduces a new 'Last
operator in do {...} must be an expression' error, which then
disappears if I explicitly return e
expr :: Parser Int
expr = do t - term
do symbol +
e - expr
return e
return (t + e)
+++ return t
to give me the original undefined variable t error at the line expr =
do t - term . It looks in scope to me... :(
Thanks,
Nik
Dr Nik Freydís Whitehead
University of Akureyri, Iceland
*
Having the moral high ground is good.
Having the moral high ground and an FGMP-15 is better.
*
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Re: [Haskell-cafe] Must be a FAQ - Can't make install Hugs98-Nov2003 on MacOSX 10.3.8
See the hugs-bugs archive: http://www.mail-archive.com/hugs-bugs@haskell.org/msg02815.html Malcolm Wallace wrote: The configure script is (wrongly) determining that the MacOS X C compiler does not support Floats/Doubles. Ideally, the autoconf magic which determined this setting should be fixed, but in the meantime, you can work around it by: (1) Edit src/config.h, replacing the line /* #undef HAVE_LIBM */ with #define HAVE_LIBM 1 (2) rm src/*.o (3) cd src; make Regards, Malcolm Hope this answers your question. Arthur On 25-feb-05, at 16:09, Kaoru Hosokawa wrote: I'm sure this must be a FAQ, but I can't make install hugs98-Nov2003 on MacOS 10.3.8. The error message I get is: ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Problem with type signature in local function
You could also do it with the current implementation of scoped type
variables. So you don't need to compile the CVS HEAD version. Note that
you need the pattern (n1::a) and/or (n2::a) to introduce the scoped
type variable, which was not necessary in Ralf's solution.
{-# OPTIONS -fglasgow-exts #-}
add :: Num a = a - a - a
add (n1::a) n2 = addToN1 n2
where addToN1 :: a - a
addToN1 number = n1 + number
Cheers,
Arthur
On 24-dec-04, at 13:04, Ralf Laemmel wrote:
Timely question. :-)
GHC supports lexically scoped type variables.
And this support has just been _extended_ to be readily useful for you
(see discussion on the ghc list just a few days back).
With GHC CVS HEAD the following program works fine.
(Note that I removed the inner Num constraint.)
{-# OPTIONS -fglasgow-exts #-}
add :: Num a = a - a - a
add n1 n2 = addToN1 n2
where addToN1 :: a - a
addToN1 number = n1 + number
... even though I would not claim that this is a really strong example
of lexically scoped variables, but you probably just simplied a real
code example.
And most of my recent code I use lexically scoped type variables
extensively. They are great.
(In principle, one does not need them at the cost of encoding. Using
asTypeOf and friends ...)
Merry Xmas,
Ralf
Christian Hofer wrote:
Hi,
in order to understand the reason for strange compiler error
messages, I often find it convenient to add type signatures to every
function. But sometimes this seems to be impossible when using type
variables.
I try to give a very easy example:
add :: Num a = a - a - a
add n1 n2 = addToN1 n2
where addToN1 :: Num a = a - a
addToN1 number = n1 + number
ghc says Inferred type is less polymorphic than expected, and
justly so, because the type variable a in addToN1 must of course be
the same as in add. Is there a way to specify this?
Regards,
Chris
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Re: [Haskell-cafe] Re: generalised algebraic data types, existential types, and phantom types
You could make use of the equality data type[1,2]:
newtype Equal a b = Eq (forall f . f a - f b)
cast :: Equal a b - (a - b)
cast (Equal f) = f id
Your data type would be:
data Term a = Lit Int (Equal Int a)
| Inc (Term Int) (Equal Int a)
| ...
| forall b c . Pair (Term b) (Term c) (Equal (b,c) a)
eval :: Term a - a
eval (Lit x eq) = cast x eq
eval ... = ...
A value of type 'Equal a b' serves as a proof that 'a' and 'b' are
equal. Its use is described in [1] and [2]. Pasalic [3] makes use of
this type to implement typed abstract syntax. This paper describes what
you are trying to achieve with your 'Term' data type.
Arthur
[1]Typing Dynamic Typing, Arthur Baars and Doaitse Swierstra. ICFP 2002.
http://www.cs.uu.nl/~arthurb/dynamic.html
[2]A lightweight implementation of generics and dynamics, James Cheney
and Ralf Hinze. Haskell Workshop 2002, Pittsburgh, PA, 2002.
http://www.cs.cornell.edu/people/jcheney/papers/Dynamic-final.pdf
[3]Emir Pasalic: Meta-Programming with Typed Object-Language
Representations.
http://www.cse.ogi.edu/~pasalic/
On 22-jul-04, at 18:06, Jim Apple wrote:
I tried to post this to gmane.comp.lang.haskell.general, but it never
got there - it may belong here anyway.
Abraham Egnor wrote:
Is there any closer approximation [of GADTs] possible?
{-# OPTIONS -fglasgow-exts #-}
{-# OPTIONS -fallow-undecidable-instances #-}
data Term a = forall b . (EqType a b, EqType b Int) = Lit Int
| forall b . (EqType a b, EqType b Int) = Inc (Term Int)
| forall b . (EqType a b, EqType b Bool) = IsZ (Term Int)
| If (Term Bool) (Term a) (Term a)
| forall b . Fst (Term (a,b))
| forall b . Snd (Term (b,a))
| forall b c . (EqType a (b,c))= Pair (Term b) (Term c)
class EqType a b | a-b, b-a
instance EqType a a
Unfortunately, I can't seem to write an eval function:
eval (Lit a) = a
gives
Inferred type is less polymorphic than expected
Quantified type variable `b' is unified with `Int'
When checking an existential match that binds
The pattern(s) have type(s): Term Int
The body has type: Int
In the definition of `eval': eval (Lit x) = x
Any ideas?
Jim
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Re: [Haskell-cafe] What's wrong with my Haskell code?
It is because you use 'div' instead of '/'.
div :: Integral a = a - a - a
(/) :: Fractional a = a - a - a
Rationals are instance of the class Fractional, but not of Integral
Prelude :i Fractional
class Num a = Fractional a where
(/) :: a - a - a
recip :: a - a
fromRational :: Rational - a
fromDouble :: Double - a
-- instances:
instance Fractional Float
instance Fractional Double
instance Integral a = Fractional (Ratio a)
Hope this helps,
Arthur
On 13-apr-04, at 20:44, Meihui Fan wrote:
when loading the following code, hugs escaped and reported that
ERROR cal24.hs:10 - Instance of Integral (Ratio Integer) required
for definition of Main.eval
I don't know why and how to solve it, anyone help me?
data ETree = Add ETree ETree
| Sub ETree ETree
| Mul ETree ETree
| Div ETree ETree
| Node Integer
deriving Show
eval :: ETree-Maybe Rational
eval (Node x)= Just (fromInteger x)
eval (Add t1 t2) = do { x-eval t1; y-eval t2;
if x=y then return (x+y) else Nothing }
eval (Sub t1 t2) = do { x-eval t1; y-eval t2; return (x-y) }
eval (Mul t1 t2) = do { x-eval t1; y-eval t2;
if x=y then return (x*y) else Nothing }
eval (Div t1 t2) = do { x-eval t1; y-eval t2;
if y/=0 then return (x `div` y) else Nothing }
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Re: lexer puzzle
I agree with Marcin, A... should be split into A.. and . As I read the (on-line) report the maximal munch rule says that you should read the longest lexeme. It does not say that two operators have to be separated by whitespace. Because A... is not a lexeme, the longest lexeme you can read from A... is A.. (qualified dot-operator). Arthur On maandag, sep 15, 2003, at 12:11 Europe/Amsterdam, Malcolm Wallace wrote: Marcin 'Qrczak' Kowalczyk [EMAIL PROTECTED] writes: Argh, I was wrong. It's A.. (qualified operator), then . (operator). You are forgetting about the maximal munch rule. An operator cannot appear directly next to another operator without some whitespace to separate them. For instance A.+. is an operator called (+.) from module A, not an operator called + followed by compose. But, although A could be the three-dot operator ... from the module A, it is not possible to have A... interpreted as a two-dot operator, because .. is reserved as sugar for enumeration sequences, and so is explicitly excluded from the varsym production. Thus, the only possible lexical interpretation is the one you first suggested, namely a constructor A followed by a three-dot operator Regards, Malcolm ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Announce: wxhaskell: MacOSX installer
I have created an out-of-the-box MacOS X installer for wxHaskell. Simply double-click on the package. The installer can be downloaded from: http://home.deds.nl/~dediav/WxHaskell.dmg This link is only valid for a couple of weeks. After Daan returns from holidays he'll probably put it on the wxHaskell pages. Note the package only works with ghc 6.0 or higher! The installer does the following: 1) copy wxWindows and wxHaskell libraries to /usr/local/wxhaskell 2) register the wx and wxh packages with ghc-pkg To uninstall, type in a Terminal: 1) rm /usr/local/wxhaskell 2) ghc-pkg -r wxh 3) ghc-pkg -r wx The package also contains the wxHaskell documentation and samples. Copy them to your disk. To test the installation: Open a terminal in the samples/wx directory, and write: 1) ghc --make -package wx Camels.hs -o camels 2) /usr/local/wxhaskell/bin/macosx-app --program=camels 3) ./camels The macosx-app script is located at /usr/local/wxhaskell/bin; you may want to add this directory to your PATH. Have fun :-)) Arthur From: Daan Leijen [EMAIL PROTECTED] Date: din jul 22, 2003 19:47:27 Europe/Amsterdam To: [EMAIL PROTECTED], [EMAIL PROTECTED], [EMAIL PROTECTED] Subject: [wxhaskell-users] Announce: wxhaskell: macOSX binaries and bugfix. Reply-To: [EMAIL PROTECTED] Two announcements regarding wxHaskell: http://wxhaskell.sourceforge.net 1) I have created a binary release for MacOS X. Installation details can be found on the download page. 2) I have fixed a small bug in the configuration script for Linux. The source release has been updated. (and I promise I won't do that again as I go tomorrow on a holiday for two weeks :-) All the best, Daan. _ Tired of spam? Get advanced junk mail protection with MSN 8. http://join.msn.com/?page=features/junkmail ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
GHC 6 for Mac OS X: fix for package util/readline
On Mac OS X GHCi 6.0 complains about a missing symbol when trying to load package util. Wolgang Thaller told me the cause of this bug: The problem was that there was a file called HsReadline.c. The corresponding object file, HsReadline.o, replaced HSreadline.o on the Mac's case-insensitive file system. So the actual problem is a broken readline package, which is used by util. The bug is fixed in the next release (6.0.1) and in CVS. For those who don't feel like compiling CVS-sources, or waiting for the next release; below is a quick fix. A working HSreadline.o is created from the libHSreadline.a file using the -g option of ghc-pkg. 1) set environment variable libdir to ghc-6's directory: setenv libdir /usr/local/lib/ghc-6.0/ 2) cp $libdir/libHSreadline.a $libdir/libHSreadline_p.a /tmp/ 3) create a config file for package readline: ghc-pkg -s readline /tmp/readline.conf 4) cd /tmp 5) remove the package readline from ghc: ghc-pkg -r readline 6) add package again and let ghc-pkg create the ghci library: ghc-pkg -g -a -i /tmp/readline.conf For step 5/6 you may need root-access, use su or sudo Arthur ___ Glasgow-haskell-bugs mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
Re: clueless GHCI user wishes to load QuickCheck
QuickCheck is in de util package. You can load a package with the
-package flag:
$ ghci -package util
Prelude :browse QuickCheck
class Arbitrary a where {
arbitrary :: Gen a; coarbitrary :: forall b. a - Gen b - Gen b; }
arbitrary :: forall a. (Arbitrary a) = Gen a
...
Prelude:module QuickCheck
Prelude QuickCheck :info trivial
-- trivial is a variable
trivial :: forall a. (Testable a) = Bool - a - Property
Hope this helps.
Cheers,
Arthur
On Tuesday, March 11, 2003, at 07:02 PM, Norman Ramsey wrote:
Can anyone help me figure out how to load QuickCheck into GHCI?
QuickCheck is included in my Debian package, but my attempts
at loading it are bootless:
Prelude :load QuickCheck
can't find module `QuickCheck'
Prelude :load util/QuickCheck
can't find module `util/QuickCheck'
Prelude :info
syntax: `:i thing-you-want-info-about'
Prelude :load util/QuickCheck.hi
can't find module `util/QuickCheck.hi'
Prelude :load /usr/lib/ghc-5.02.2/imports/util/QuickCheck.hi
can't find module `/usr/lib/ghc-5.02.2/imports/util/QuickCheck.hi'
Prelude :load /usr/lib/ghc-5.02.2/imports/util/QuickCheck
can't find module `/usr/lib/ghc-5.02.2/imports/util/QuickCheck'
Prelude
Any advice, anyone?
Norman
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Building snapshot ghc-5.05.20030123 on macOSX
I want to play a bit with Template Haskell, so I downloaded a cvs snapshot (ghc-5.05.20030123.tar.gz) and tried to compile it on my mac (OSX 10.2). I figured out that I needed a stage2 or stage3 compiler to get TH working, so I tried 'make stage3' . GHC started to compile itself three times. At lot of hours later it finished. Unfortunately, when I tried run the thing on a Prinf example, I got the following message. compiler/stage3- ./ghc-inplace -c Printf.hs -fglasgow-exts -package haskell-src Template Haskell bracket illegal in a stage-1 compiler [| \s - s |] The thing incompiler/stage3 was just a stage-1 compiler, so ghc had achieved nothing useful after re-compiling three times :-( The problem is that GHCi support is not turned on for MacOSX. I guess GHCi is used to do compile-time evaluation of code for TH. In file mk/config.mk.in it says: # Include GHCi in the compiler. Default to NO for the time being. ifneq $(findstring $(HostOS_CPP), mingw32 cygwin32 linux solaris2 freebsd netbsd openbsd) GhcWithInterpreter=YES else GhcWithInterpreter=NO endif In the list of supported systems MacOSX is missing. After adding darwin to this list, and waiting for hours, I got a working TH-compiler. Can someone add OSX(darwin) to the mk/config.mk.in file? It would also be nice to get a warning if one tries to compile a stage-(N1) compiler without GHCi support. Cheers, Arthur ___ Glasgow-haskell-users mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: question about parsing integers and floats with Parsec
Hi,
Here is a solution that does not use try.
import Parsec
import qualified ParsecToken as P
import ParsecLanguage (emptyDef)
lexer = P.makeTokenParser emptyDef
-- naturalOrFloat parses a float or an integer without sign
naturalOrFloat = P.naturalOrFloat lexer
data Sign = Positive | Negative
applySign :: Num a = Sign - a - a
applySign Positive = id
applySign Negative = negate
sign :: Parser Sign
sign = do { char '-'
; return Negative
}
| do { char '+'
; return Positive
}
| return Positive
genericgrab :: Parser (Either Integer Double)
genericgrab = do { s - sign
; num - naturalOrFloat
; return (case num of
Right x - Right (applySign s x)
Left x - Left (applySign s x)
)
}
On 16/8/02 7:08, Harris, Andrew [EMAIL PROTECTED] wrote:
Hi -
After a bit of monkeying around, it seems the following parser works
to detect both integers and floats, both positive and negative. I'm sure
you wizards have many better ways of doing this. Anyway here it is for your
amusement:
import Parsec
import qualified ParsecToken as P
import ParsecLanguage (emptyDef)
lexer = P.makeTokenParser emptyDef
integer = P.integer lexer
float = P.float lexer
genericgrab :: Parser (Either Integer Double)
genericgrab = try ( do { n - char '-'
; f - float
; return (Right (-f))
}
)
| try ( do { f - float
; return (Right f)
}
)
| try ( do { f - integer
; return (Left f)
}
)
-andrew
-Original Message-
From: Harris, Andrew [mailto:[EMAIL PROTECTED]]
Sent: Thursday, August 15, 2002 6:55 PM
To: '[EMAIL PROTECTED]'
Subject: question about parsing integers and floats with Parsec
Hi -
This isn't a pure Haskell question, but I'm trying to use the
Parsec library to parse out space separated numbers, which
could be integers
or floats and either positive or negative. I was using the
naturalOrFloat
lexeme parser from the ParsecToken module, until I realized
that it doesn't
seem to handle negative integers. I've been poking with
trying to pair the
integer and float lexeme parsers with try() blocks, but I
ain't no parsing
expert and am not making good progress.
Any help/hints would be appreciated!
thanks,
-andrew
---
Andrew Harris[EMAIL PROTECTED]
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Existentials and newtype
At [EMAIL PROTECTED] I asked the following:
this works
data Exists f = forall x . E ( f x)
this doesn't work
newtype Exists f = forall x . E ( f x)
Hugs accepts both.
It there a reason why existential quantification does not work with a
newtype, or is it just
a parser problem?
This problem is clearly not a bug but a restriction imposed by the designers
of GHC,
as Keith Wansbrough pointed out to me:
From the manual, section 7.12.3 Restrictions:
--8--
You can't use existential quantification for newtype declarations. So this
is illegal:
newtype T = forall a. Ord a = MkT a
Reason: a value of type T must be represented as a pair of a
dictionary for Ord t and a value of type t. That contradicts the idea
that newtype should have no concrete representation. You can get just
the same efficiency and effect by using data instead of newtype. If
there is no overloading involved, then there is more of a case for
allowing an existentially-quantified newtype, because the data because
the data version does carry an implementation cost, but single-field
existentially quantified constructors aren't much use. So the simple
restriction (no existential stuff on newtype) stands, unless there are
convincing reasons to change it.
--8--
I perfectly understand why the use of class constraints in combination with
newtype and existential quantification is
illegal. In my opinion the restriction should only forbit this particular
combination, instead of all uses of newtype with existentials.
For me single-field existentially quantified constructors are quite useful:
For example
newtype Exists f = forall x . E (f x)
data Object a = Object {value :: a , toString:: a - String }
data Thing a = Thing {val :: a , next :: a - a, run :: a - IO ()}
-- Here I hide the type variables of Object and Thing using the type Exists:
type EObject = Exists Object
type EThing= Exists Thing
without single-field existentially quantified constructors I would have to
write:
data EObject = forall a . EObject a (a - String)
data EThing = forall a . EThing a (a - a) (a- IO ())
or write the Exists type using data
When using data I have to pay a (however small) run-time cost, that I would
not have to pay
when using a newtype.
The other solution does not have this run-time cost, but it makes my code
less readable:
I have to invent new types and constructor names.
I cannot use named fields.
I cannot reuse polymorphic functions that work on Object on EObjects.
Summarizing: without single-field existentially quantified constructors in
newtypes
either my code gets less readable or I have to pay a run-time price.
Is it possible to relax the restriction on newtypes in combination with
existentials?
Cheers, Arthur
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Exists for all places
I see the difference in behaviour of the parser for newtype and data definitions in combination with universal quantifications, as reported by Ralf Hinze (forall for all places, October 2001), has been fixed in GHC version 5.02.1. I encountered a similar problem with existential quantification: this works data Exists f = forall x . E ( f x) this doesn't work newtype Exists f = forall x . E ( f x) Hugs accepts both. It there a reason why existential quantification does not work with a newtype, or is it just a parser problem? Cheers, Arthur ___ Glasgow-haskell-bugs mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
