Hdirect
Hi there. I get the impression someone is working on this stuff from the changes in CVS so you probably know about these problems, but just in case (and to prove that it is all worthwhile, for you to do this stuff), here they are: Using the latest Win32 GHC (thanks to Reuben for the ongoing work) and CVS source for HDirect: - Trying to build Hdirect straight from CVS has a number of dependency problems. I find I have to do several iterations of make and make boot and make lib. (Happy likewise has a crossdependency when bootstrapping without installing a binary copy first) - The dependency analysis seems to fail for IDLUtils.lhs - I have to do a make IDLUtils.hi - You have to build ihc.exe by bootstrapping without com first. - make clean does not delete src/ihc.exe - the com interface files seem to think they are in package main eg: __interface Main PointerPrim 1 501 where __export PointerPrim finalFreeMemory finalNoFree primAllocFrame primAllocMemory primFinalise primFreeBSTR primFreeMemory primNoFree; import Prelude :: 1; import PrelBase ! :: 1; import PrelWord :: 1; import PrelIOBase :: 1; and so at build time in the comcli example: -- miketh@NASTURTIUM //e/cvs/ghc/hdirect/examples/comcli $ make /cygdrive/e/ghc/ghc-5.01/bin/ghc -fglasgow-exts -package com -fno-warn-missing-m ethods -O-c Quartz.hs -o Quartz.o Quartz.hs:9: Module `Automation' is located in package `com' but its interface file claims it is part of package `Main' Quartz.hs:13: Module `Com' is located in package `com' but its interface file claims it is part of package `Main' make: *** [Quartz.o] Error 1 -- - various minor source problems: Index: lib/ComDll.lhs === RCS file: /cvs/fptools/hdirect/lib/ComDll.lhs,v retrieving revision 1.16 diff -r1.16 ComDll.lhs 39c39 import Foreign --- import Foreign hiding ( Ptr ) Index: lib/ComServ.lhs === RCS file: /cvs/fptools/hdirect/lib/ComServ.lhs,v retrieving revision 1.16 diff -r1.16 ComServ.lhs 88c88 import Foreign --- import Foreign hiding ( Ptr ) Index: lib/autoPrim.h === RCS file: /cvs/fptools/hdirect/lib/autoPrim.h,v retrieving revision 1.10 diff -r1.10 autoPrim.h 100c100 #ifdef __GNUC__ --- #if defined(__GNUC__) ! defined(__MINGW32__) Index: src/Makefile === RCS file: /cvs/fptools/hdirect/src/Makefile,v retrieving revision 1.57 diff -r1.57 Makefile 52c52 --- SUPPORT_TYPELIBS = YES Index: src/TLBWriter.lhs === RCS file: /cvs/fptools/hdirect/src/TLBWriter.lhs,v retrieving revision 1.23 diff -r1.23 TLBWriter.lhs 36c36 writeTLB :: [String] - [Decl] - IO () --- --writeTLB :: [String] - [Decl] - IO () 38c38 writeTLB _ _ = ioError (userError (writeTLB: type library writer code not com piled in)) --- --writeTLB _ _ = ioError (userError (writeTLB: type library writer code not c ompiled in)) Cheers (and as usual thanks for the good work) Mike Thomas ___ Glasgow-haskell-bugs mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
GHC derives and accepts illegal Haskell98 contexts
I have experimented with the definition of class assertions in section 4.1.3 in the Haskell 98 report, and found that GHC 4.08 and GHC 5.00.1 (invoked via ``ghc -c Test.hs'') both derive and accept type arguments (in class assertions) that do not have a variable as head. I cannot really imagine where this might hurt, though. By the way, I find that the .hi format of 5.00 is, from a human-readability point of view, in my opinion not optimal in that there is no separator after the __forall-bound variables: r2 :: __forall b (E (PrelBase.String, b)) = b - PrelBase.Bool; ^ There is no comparison with NHC since NHC still has the ``simple context restriction''. Best regards, Wolfram = module Test where class T a where t :: a b c - Bool class E a where e :: a - a - Bool instance T (,) where t (a,b) = True q1 :: (T a,Eq (a String b)) = a String b - Bool q1 x = if x == x then t x else True --r1 :: Eq (String, b) = b - Bool -- derived and accepted by GHC --r1 :: Eq a = a - Bool -- derived by Hugs98, -- accepted by both Hugs98 and GHC r1 x = q1 (asd,x) q2 :: (T a,E (a String b)) = a String b - Bool q2 x = if e x x then t x else True --r2 :: E (String, b) = b - Bool -- derived and accepted by GHC r2 x = q2 (asd,x) -- not accepted by hugs +98 ___ Glasgow-haskell-bugs mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-bugs
Booting from HC files is completely broken
Hi all. I tried to compile ghc-5.00.2 on a RedHat-7.1 system by booting from HC files. (Actually, I have a ghc-5.00 installed, but apparently some programs compiled with it segfault, and this makes compilation of ghc-5.00.2 impossible. Note that I have a working happy-1.10, however.) Now it seems that the distrib/hc-build script is broken. First of all, this is a minor point, but it calls ./configure twice, and the second time it omits passing $configopts to it, so that anything like --prefix will fail. But this is not the problem I have. To avoid the above gotcha, I manually drove the various commands. Up to the second invocation of configure, everything works fine. I end up with a working ghc-inplace (well, it seems to work). But no interface (.hi) files. How on earth are these supposed to be generated? The hc-build script calls ./configure a second time; now that does not find a working ghc (of course: it doesn't know about the just built ghc-inplace), so GHC= in the Makefiles, and the attempt to make all in ghc/utils fails miserably (in a ridiculous way, in fact: since GHC=, the next argument on the command line, viz. -ldl, becomes the command, and GNU make looks for a program called ldl, doesn't find it, and ignores the error because of the prepended -; similarly, some -o commands look for a program named o and ignore the error). I tried passing GHC to point to the ghc-inplace program built, but that doesn't work either (missing interface files, so ghc/utils can't be built; and ghc/libs can't be built because of the missing ghc/utils). It seems that we still have a chicken-and-egg problem which the .hc files have done nothing to solve. Any suggestion (other than starting from a binary build, which is what I suppose I'll try now)? (Note: I'm not subscribed to the list. Please Cc me all replies.) -- David A. Madore ([EMAIL PROTECTED], http://www.eleves.ens.fr:8080/home/madore/ ) ___ Glasgow-haskell-users mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
RE: Booting from HC files is completely broken
I tried to compile ghc-5.00.2 on a RedHat-7.1 system by booting from HC files. (Actually, I have a ghc-5.00 installed, but apparently some programs compiled with it segfault, and this makes compilation of ghc-5.00.2 impossible. Note that I have a working happy-1.10, however.) Now it seems that the distrib/hc-build script is broken. Yes it is, sorry about that. There are several open bug reports on SourceForge relating to the broken hc bootstrapping situation. I'm working on fixing it, so hopefully the next release will have reliable hc bootstrapping again. Your particular problem can be solved by (a) getting the .hs files that are generated from the .hsc files in the libraries, and (b) making in ghc/lib/std and hslibs before ghc/utils. Having the .hs files means you don't need hsc2hs in order to build the libraries and works around the chicken/egg problem. Cheers, Simon ___ Glasgow-haskell-users mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: hGetContents and laziness in file io
Hi, My guess is that there is a space leak in your program. In both function convert and parseAll, there are references (the variable ulf) to the contents of the input file, and they will probably not be released until the functions return (unless you use a compiler that is clever enough to delete references after their last use...). There might be sources of space leaks also in the function parse that is called from parseAll. If your program only processes one file each time you run it, you could structure it like this: main = interact parseAll' where parseAll :: String - String parseAll' = unlines . map convert' . parse' parse' :: String - [Tree] convert' :: Tree - String parse' s = case parseOneTree s of Good (tree,rest) - tree:parse' rest Error err - error err convert' tree = ... Note that parse' is lazy: it returns the first tree before it tries to parse the rest of the input. Anyway, space leaks can be hard to find and eliminate, but there are tools that can help. The Haskell compiler Nhc98 (http://www.cs.york.ac.uk/fp/nhc98/) tries to generate space efficient code to begin with, but also provides heap profiling to help you find out what kind of data is occupying all the space (constructor profile), which functions produced the data (producer profile) which functions have references to the data (retainer profile), ... Hope this helps! Thomas Hallgren Hal Daume wrote: ... the file that I'm working with is ~20mb of trees. When I run my program on this, it is unable to reclaim space (unless i set the heap really high). ... convert inF outF = do inH - openFile inF ReadMode ulf - hGetContents inH outH - openFile outF WriteMode parseAll outH ulf hClose inH hClose outH parseAll outH ulf = case parse s of Good (tree, rest) - case convert tree of Good s' - do hPutStrLn outFile s' Error err - do putStrLn err Error err - do return () PLEASE help! ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
beginner's questions - fix f
Hello, After watching members of this list discussing some for me really incomprehensible details of Haskell standard, I feel sort of silly for asking something probably very basic, but I'd be very grateful for any answer: Some time ago, I have decided to educate myself in functional programming, and I choose Haskell as a language of choice. I purchased Haskell school of expression and began to study it. Everything went well, I understood (at least intuitively) everything, but then, suddenly, I hit a wall: Exercise 9.9: remainder a b = if a b then a else remainder (a-b) b fix f = f (fix f) Rewrite remainder using fix so that it is not recursive. Function fix left me completely puzzled. With help of hugs I found out that its type is ( a - a ) - a, but I have absolutely no idea how it could be used to do anything useful. It seems to me that argument to f on the right side of fix's definition must always evaluate to the same value, no matter how deep the recursion is, so there is no point to use it. I guess I am missing something very obvious, but what is it? Can somebody provide me with an example how to use fix for something just a bit useful, if possible to rewrite remainder? thank you, Bob Koutsky ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: beginner's questions - fix f
Bob Koutsky wrote: remainder a b = if a b then a else remainder (a-b) b fix f = f (fix f) Rewrite remainder using fix so that it is not recursive. Function fix left me completely puzzled. With help of hugs I found out that its type is ( a - a ) - a, but I have absolutely no idea how it could be used to do anything useful. Function fix is a so-called fixpoint operator. Theory says that you can formulate any computable function using only non-recursive definitions plus fix. Can somebody provide me with an example how to use fix for something just a bit useful, if possible to rewrite remainder? Try: remainderF self a b = if a b then a else self (a-b) b remainder = fix remainderF From this example it is easy to infer how to transform arbitrary recursive definitions. Even generalising it to mutual recursion is not difficult (and left as an exercise to the reader ;-). All the best, - Andreas -- Andreas Rossberg, [EMAIL PROTECTED] Computer games don't affect kids. If Pac Man affected us as kids, we would all be running around in darkened rooms, munching pills, and listening to repetitive music. ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: lexical description problem in language report?
Thomas Hallgren wrote There seems to be a similar problem with qualified identifiers. The production for lexeme includes varid, conid, etc, rather than qvarid, qconid, etc. (Perhaps someone forgot to update it when qualified names were introduced, in Haskell 1.3...) Sorry we must have a different version of the report, but in my copy and also in the version on Simon PJ's web-pages section 2.4 (in the last paragraph) and appendix B include productions for the qualified identifiers. Wolfgang -- Wolfgang Lux Phone: +49-251-83-38263 Institut fuer Wirtschaftinformatik FAX: +49-251-83-38259 Universitaet Muenster Email: [EMAIL PROTECTED] ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: Picky details about Unicode (was RE: Haskell 98 Report possible errors, part one)
Mon, 23 Jul 2001 11:23:30 -0700, Mark P Jones [EMAIL PROTECTED] pisze: I guess the intention here is that: symbol - ascSymbol | uniSymbol_special | _ | : | | ' Right. In fact, since all the characters in ascSymbol are either punctuation or symbols in Unicode, the inclusion of ascSymbol is redundant, and a better specification might be: symbol - uniSymbol_special | _ | : | | ' It would still be nice to explicitly list ASCII symbols, so one doesn't need to look at Unicode specs to use ASCII-only source. There are two places when character predicates are used in Haskell: program source and module Char. I'm sure that we all agree that they should be consistent with each other. Some predicates in module Char are wrong, i.e. I don't agree with their meaning. For example that isSpace is restricted to ISO-8859-1, and that caseless letters are considered uppercase. It's not clear what good definitions are, or even what set of predicates is useful, because there is no single official source with unambiguous and complete set of predicates. There are Unicode character categories, Unicode property lists, and implementations of C character predicates - all with different data. I guess Java specs have something to tell here too. I have an implemented proposal of improved Char predicates in QForeign http://sf.net/projects/qforeign/. Definitions are based on both Unicode character categories and PropList.txt from Unicode. -- __( Marcin Kowalczyk * [EMAIL PROTECTED] http://qrczak.ids.net.pl/ \__/ ^^ SYGNATURA ZASTÊPCZA QRCZAK ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: beginner's questions - fix f
From: Bob Koutsky [EMAIL PROTECTED] Date: Tue, 24 Jul 2001 09:49:33 +0200 [...] suddenly, I hit a wall: Exercise 9.9: remainder a b = if a b then a else remainder (a-b) b fix f = f (fix f) Rewrite remainder using fix so that it is not recursive. Function fix left me completely puzzled. With help of hugs I found out that its type is ( a - a ) - a, but I have absolutely no idea how it could be used to do anything useful. It seems to me that argument to f on the right side of fix's definition must always evaluate to the same value, no matter how deep the recursion is, so there is no point to use it. I guess I am missing something very obvious, but what is it? Can somebody provide me with an example how to use fix for something just a bit useful, if possible to rewrite remainder? Well, it is a kind of sleight of hand. The idea is that if you already had a working remainder function, you could use that to write another, like this: remainder2 a b = if a b then a else remainder (a-b) b That's not very interesting. But if you made the 'working' function into an argument of a 'step' function: rem_step f a b = if a b then a else f (a-b) b you could actually define remainder like this: remainder3 = rem_step remainder3 Now this is still a recursive definition, but it makes it easier to see how the fix function works in a minute. An example evaluation: remainder3 3 2 == rem_step remainder3 3 2 == if 3 2 then 3 else remainder3 (3-2) 2 == remainder3 1 2 == rem_step remainder3 1 2 == if 1 2 then 1 else remainder3 (1-2) 2 == 1 Now, anything that's defined as x = f x is called a fixpoint of f. It's possible to prove that there's only one (when f is a Haskell function, at least) so we can talk of 'the' fixpoint. Since we can rewrite all recursive definitions in this form, we could get rid of all explicit recursion if we had a 'fixpoint operator' that would solve the equation for us. That operator is usually called 'fix', and since we want fix f to be a solution to the equation, we get the defining equation in your exercise: fix f = f (fix f) This looks like a recursive definition again, but if you want a system without explicit recursion you have to disallow this as a function. In such a system, fix is a builtin operator, and the equation can be seen as a rewrite rule that lets you reason about the evaluation of expressions that contain it. However, it turns out that Haskell is quite happy to use the definition as it stands. Let's redo the example: fix f = f (fix f) remainder4 = fix rem_step remainder4 3 2 == fix rem_step 3 2 == rem_step (fix rem_step) 3 2 == if 3 2 then 3 else (fix rem_step) (3-2) 2 == fix rem_step 1 2 == rem_step (fix rem_step) 1 2 == if 1 2 then 1 else (fix rem_step) (1-2) 2 == 1 As you noted, each occurence of fix rem_step above is in a sense the same, and could evaluate in the same way. But since the expression is a function (in this case), how it actually behaves depends on its arguments --- I think that's really the point you were looking for: because Haskell is lazy, and there's no structure sharing between the occurences, each occurence is only expanded as often as necessary. Note that operationally, this Haskell version of the fixpoint operator is only sort-of equivalent to a Haskell recursive definition (which is defined in terms of a 'real' fixpoint). That's even clearer with data structures; compare zeroes = 0 :: zeroes and zer_step l = 0 :: l zeroes2 = fix zer_step zeroes is a circular list with one element, while zeroes2 is an infinite list with all zero elements. It may not be easy to tell the difference from inside a Haskell program, but the sound of the disk thrashing will sometimes alert the programmer to it. Lars Mathiesen (U of Copenhagen CS Dep) [EMAIL PROTECTED] (Humour NOT marked) ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: Haskell 98 Report possible errors, part one
From: Dylan Thurston [EMAIL PROTECTED] Date: Mon, 23 Jul 2001 19:57:54 -0400 On Mon, Jul 23, 2001 at 06:30:30AM -0700, Simon Peyton-Jones wrote: Someone else, quoted by Simon, attribution elided by Dylan, wrote: | 2.2. Identifiers can use small and large Unicode letters. | What about caseless scripts where letters are neither small | nor large? The description of module Char says: For the | purposes of Haskell, any alphabetic character which is not | lower case is treated as upper case (Unicode actually has | three cases: upper, lower and title). This suggests that the | only anomaly is that titlecase letters are considered | uppercase. But what is actually specified is that caseless | scripts can be used to write constructor names, but not to | variable names. I don't know how to solve this. I am woefully ignorant of Unicode, and I have no idea what to do about this one. I therefore propose to do nothing on the grounds that I might easily make matters worse. In this case, what about requiring identifiers to start with an upper or lower case alphabetic character? I'm not sure that makes things better. It just makes it impossible to have identifiers in caseless scripts (some of which are alphabetic). And whether you choose your upper or lower case alphabetic character from Latin, Greek, Coptic, Cyrillic, Armenian, Georgian, or Deseret, it will probably look silly in front of a variable name spelled in Hangul. What would make sense to me is to define that caseless letters (Unicode class Lo) behave as lowercase, and to choose some easily visible, culturally neutral, symbol as the official 'conid marker'. Since the problem only arises on Unicode-capable systems, there should be plenty of those to choose from, even outside Latin-1. To fix Haskell 98, the least intrusive way might be to allow only classes Ll, Lt, and Lu in identifiers, with Lt (titlecase) and Lu counting as uppercase --- it looks like that may actually have been the intention. And then add a note explaining that caseless scripts can't be used because they weren't considered initially. Lars Mathiesen (U of Copenhagen CS Dep) [EMAIL PROTECTED] (Humour NOT marked) ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: beginner's questions - fix f
24 Jul 2001 12:04:33 -, Lars Henrik Mathiesen [EMAIL PROTECTED] pisze: Now, anything that's defined as x = f x is called a fixpoint of f. It's possible to prove that there's only one (when f is a Haskell function, at least) so we can talk of 'the' fixpoint. Not necessarily only one, e.g. any value is a fixpoint of identity function. But there is one *least* fixpoint wrt. definedness, and it can be effectively found. If the function is strict, the least fixpoint is not very interesting because it's bottom, a value representing nontermination or error. If the function is not strict, bottom is not its fixpoint, so the fixpoint obtained by x = f x or fix f is more interesting. BTW, a better definition than fix f = f (fix f) is fix f = let x = f x in x because it increases sharing, avoiding recomputation. fix f = f (fix f) This looks like a recursive definition again, but if you want a system without explicit recursion you have to disallow this as a function. In such a system, fix is a builtin operator, You can define fix without recursion in a typeless lambda calculus: fix f = (\x - f (x x)) (\x - f (x x)) and this can even be stretched to fit Haskell's type system by smart use of algebraic types. Note that operationally, this Haskell version of the fixpoint operator is only sort-of equivalent to a Haskell recursive definition (which is defined in terms of a 'real' fixpoint). That's even clearer with data structures; compare zeroes = 0 :: zeroes and zer_step l = 0 :: l zeroes2 = fix zer_step zeroes is a circular list with one element, while zeroes2 is an infinite list with all zero elements. This is the difference between two definitions of fix above. -- __( Marcin Kowalczyk * [EMAIL PROTECTED] http://qrczak.ids.net.pl/ \__/ ^^ SYGNATURA ZASTÊPCZA QRCZAK ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: beginner's questions - fix f
From: Marcin 'Qrczak' Kowalczyk [EMAIL PROTECTED] Date: 24 Jul 2001 13:05:25 GMT 24 Jul 2001 12:04:33 -, Lars Henrik Mathiesen [EMAIL PROTECTED] pisze: Now, anything that's defined as x = f x is called a fixpoint of f. It's possible to prove that there's only one (when f is a Haskell function, at least) so we can talk of 'the' fixpoint. Not necessarily only one, e.g. any value is a fixpoint of identity function. But there is one *least* fixpoint wrt. definedness, and it can be effectively found. My error. I meant to write only one relevant fixpoint. This was a beginner's question, after all, so I wasn't going to go into too many details. BTW, a better definition than fix f = f (fix f) is fix f = let x = f x in x because it increases sharing, avoiding recomputation. And it very obviously constructs a solution to x = f x and delivers it. But trying to explain its operational behaviour gets right back to Haskell's builtin recursion, which wasn't what the original poster was trying to understand. You can define fix without recursion in a typeless lambda calculus: fix f = (\x - f (x x)) (\x - f (x x)) and this can even be stretched to fit Haskell's type system by smart use of algebraic types. I know. I even posted about it to another thread about two weeks ago. But that loses sharing again, of course. Lars Mathiesen (U of Copenhagen CS Dep) [EMAIL PROTECTED] (Humour NOT marked) ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Minor inconsistency in the report wrt to qualified names
In section 5.5.1 the report states that a top-level declaration brings into scope qualified and unqualified names. Thus, the following declaration is legal: module Foo where ones = 1 : Foo.ones However, in section 2.4 the report says External names may optionally be qualified in certain circumstances by prepending them with a module identifier. And also in section 3.2 the report says Qualified names may only be used to reference and imported variable or constructor, from which the above program would not be legal (because Foo.ones is not an imported constructor or variable). In order to make the report consistent, I would suggest to change the section 3.2 to say: Qualified names may be used to reference variables or constructors which are imported from other modules or defined at the top level. In section 2.4, the word External should be dropped from the sentence. Wolfgang -- Wolfgang Lux Phone: +49-251-83-38263 Institut fuer Wirtschaftinformatik FAX: +49-251-83-38259 Universitaet Muenster Email: [EMAIL PROTECTED] ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: lexical description problem in language report?
Wolfgang Lux wrote:
Thomas Hallgren wrote
There seems to be a similar problem with qualified identifiers. The
production for lexeme includes varid, conid, etc, rather than qvarid,
qconid, etc.
Sorry we must have a different version of the report, but in my copy
and also in the version on Simon PJ's web-pages section 2.4 (in the
last paragraph) and appendix B include productions for the qualified
identifiers.
The version of the report I looked in is here:
http://research.microsoft.com/Users/simonpj/haskell98-revised/haskell98-report-html/lexemes.html#sect2.2
which is probably the same one you have, so it seems I need to clarify
what I wrote.
Although qualified names are listed in section 2.4,and in appendix B,
the two first productions of the grammar are:
program - {lexeme | whitespace }
lexeme - varid | conid | varsym | consym | literal | special |
reservedop | reservedid
There is no reference to qualified names here. I thought the purpose of
these productions were to say that a Haskell program is correct on the
lexical level iff there is a derivation of it in the lexical grammar,
starting from the nonterminal program. Since qualified names are not
part of this grammar, they are not part of the lexical syntax, which
contradicts the text in section 5.5.1.
So, I repeat my improvment suggestions: include qvarid, qconid, etc, in
the production for lexeme. Move the explanation of the lexical
properties of qualified names from section 5.5.1 to section 2.4.
Regards,
Thomas Hallgren
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Re: beginner's questions - fix f
Marcin 'Qrczak' Kowalczyk wrote: BTW, a better definition than fix f = f (fix f) is fix f = let x = f x in x because it increases sharing, avoiding recomputation. The latter definition is more likely to give you sharing, but Haskell gives you no such guarantees. There are also implementations that give you sharing for the first definition. You can define fix without recursion in a typeless lambda calculus: fix f = (\x - f (x x)) (\x - f (x x)) and this can even be stretched to fit Haskell's type system by smart use of algebraic types. But those algebraic types are recursive! -- Lennart ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: hGetContents and laziness in file io
Okay, I understand the problem. I would do something like the solution you propose, except that in the input file, trees span multiple lines. So the input file looks something like: (:cat S :subs ((() (:cat NP :subs ((() (:surf John) (() (:cat VP :subs ((() (:surf bought)) (() (:cat NP :subs ((() (:surf two)) (() (:surf cars (() (:surf . and then there will be another one directly following it (i detect the breaks because each one is completely surrounded in parens). So a line-by-line parse solution won't really do the job... ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: lexical description problem in language report?
Thomas Hallgren wrote:
program - {lexeme | whitespace }
lexeme - varid | conid | varsym | consym | literal | special |
reservedop | reservedid
There is no reference to qualified names here. I thought the purpose of
these productions were to say that a Haskell program is correct on the
lexical level iff there is a derivation of it in the lexical grammar,
starting from the nonterminal program. Since qualified names are not
part of this grammar, they are not part of the lexical syntax, which
contradicts the text in section 5.5.1.
So, I repeat my improvment suggestions: include qvarid, qconid, etc, in
the production for lexeme. Move the explanation of the lexical
properties of qualified names from section 5.5.1 to section 2.4.
You could still parse a qualified name as three lexemes.
Of course you don't want this, as this would allow white space
between them.
For the same reason, you want backquoted functions and constructors
to be only one lexeme. In order to achieve this, just use qop instead
of qvarsym and qconsym. And we need opencom, as the report says {- is
a lexeme.
So I suggest:
lexeme - qvarid | qconid | qop
| literal | special | reservedop | reservedid | opencom
It's all not new. See:
http://www.dcs.gla.ac.uk/mail-www/haskell/msg01596.html
http://www.dcs.gla.ac.uk/mail-www/haskell/msg01730.html
All the best
Christian Sievers
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newbie syntax question
Hi, I am extremely new to Haskell. This will be my first question, so go easy. I have just read Chapter 1 on Simon Thompson's book. for example a function declaration is given as follows scale : : Picture - Int - Picture If the first two types are input variables why does the syntax require me to use arrows twice? I mean isn't the following syntax more readable (hypothetically)? scale : : Picture , Int - Picture Is there a specific reason not to be able to distinguish the input parameters from the output parameter? Thanks
RE: newbie syntax question
Actually, all functions in Haskell take only one argument, although the people writing the program usually don't think of it this way. scale could alternatively have been defined with type scale :: (Picture, Int) - Picture which looks more like what we would expect in a function of two arguments. But even here there is actually only a single argument, which happens to be a pair of values. The given declaration scale :: Picture - Int - Picture is written in 'curried' form. This means that scale is a function of oneargument, 'Picture', and that its return value is a new function with type Int - Picture. That new function can immediately be applied to an int value so that it appears as if you called scale with two values instead of just one. instead of writing scale pic 3 for example, you could have written (scale pic) 3 or even let x = scale pic in x 3 Which might make it a little easier to see what is actually happening. Hope this helps, -- Gary -Original Message-From: Cagdas Ozgenc [mailto:[EMAIL PROTECTED]]Sent: Tuesday, July 24, 2001 1:07 PMTo: [EMAIL PROTECTED]Subject: newbie syntax question Hi, I am extremely new to Haskell. This will be my first question, so go easy. I have just read Chapter 1 on Simon Thompson's book. for example a function declaration is given as follows scale : : Picture - Int - Picture If the first two types are input variables why does the syntax require me to use arrows twice? I mean isn't the following syntax more readable (hypothetically)? scale : : Picture , Int - Picture Is there a specific reason not to be able to distinguish the input parameters from the output parameter? Thanks
Re: newbie syntax question
At 3:07 PM -0500 7/24/01, Cagdas Ozgenc wrote: Hi, I am extremely new to Haskell. This will be my first question, so go easy. I have just read Chapter 1 on Simon Thompson's book. for example a function declaration is given as follows scale : : Picture - Int - Picture If the first two types are input variables why does the syntax require me to use arrows twice? I mean isn't the following syntax more readable (hypothetically)? scale : : Picture , Int - Picture Is there a specific reason not to be able to distinguish the input parameters from the output parameter? Thanks Welcome to Haskell! May you find it as rewarding as I have. As you read further, you'll find that Haskell allows functions to be applied to less than their full complement of arguments. For example, if pict :: Picture, then scale pict is a function which, applied to an Int, produces a Picture. That is, scale pict :: Int - Picture This feature, known as Currying, is very useful for making rather general functions which can be specialized by supplying some of their arguments. The type notation has been designed to accommodate Currying in two ways: 0. each parameter type is followed by an arrow 1. the arrow has been designed to be right-associative. The arrow's right-associativity means that these two types are identical: scale : : Picture - Int - Picture scale : : Picture - (Int - Picture) That is, you are free to think of scale as either of a function of two arguments which returns a Picture a function of one argument which returns (a function of one argument which returns a Picture) That equivalence would be a bit less clear using the comma syntax you suggest. Cheers, --Ham -- Hamilton Richards, PhD Department of Computer Sciences Senior Lecturer Mail Code C0500 512-471-9525 The University of Texas at Austin Taylor Hall 5.138Austin, Texas 78712-1188 [EMAIL PROTECTED][EMAIL PROTECTED] -- ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: newbie syntax question
this should be on a list of the 10 first questions someone will ask when learning haskell. I have introduced several friends to the language and they all seem to ask the same things, if not always in the same order.. whats the deal with Int - Int - Int ... (currying) what does $ mean? is it 'special' syntax? (no, just a $ b = a b) why can't a - getChar be a = getChar in a do expression? how do I convert an IO a to just an a? (unsafePerformIO, but that is the wrong answer to the wrong question) general confusion about namespaces, the difference between (a,b) the type and (a,b) the value for instance. how to write some basic idioms only used in an imperitive setting, incrementing a counter, processing input, maintaining state with variables which don't directly translate to haskell usually leading to some confusion when you try to explain why they need to restructure their program. and on a more abstract level, the view of 'do' as a hack for IO rather than monads as an independant and useful abstraction. (i find the Maybe monad one of the better ways to introduce people to non-IO monads, especially if they are used to Maybe in non-monadic usage.) I have some email exchanges somewhere where I explained some of these concepts, perhaps I will edit them and add them to the Wiki, or better yet, find the best explanations from the list and add them... and of course any other newbie questions other people find they encounter... this is somewhat odd ground, because when helping someone learn the language, I WANT people to come to these questions on their own and ask them since it means they are thinking and noticing the places where they need to readjust, so its maybe not a newbies FAQ, for someone evaluating whether to learn the language, but more of a two weeks in and not quite getting everything FAQ. John -- --- John Meacham - California Institute of Technology, Alum. - [EMAIL PROTECTED] --- ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: beginner's questions - fix f
Function fix is a so-called fixpoint operator. Theory says that you can formulate any computable function using only non-recursive definitions plus fix. Could someone point me toward a proof of this? Furthermore, can any computable function be expressed in this form: fix u where u is some non-recursive term? Sam Moelius ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Machine bit representations
Hello, Is it possible in Haskell to access the underlying machine bit representation of a Float or Double value? I need to be able to be able to send this bit representation as a list of bytes in network byte order to a process running on a different platform (with a different host byte order to my platform). For reference, I run Haskell under Linux on Intel. The processes I want to communicate with run under Sun Solaris, Hitachi HPUX and Java everywhere. Any suggestions, for any Haskell translator much appreciated. Thanks. _ Global Virtual Desktop http://www.magicaldesk.com ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: Machine bit representations
At 2001-07-24 16:51, [EMAIL PROTECTED] wrote: Is it possible in Haskell to access the underlying machine bit representation of a Float or Double value? If you mean the _actual_ bit representation, then I hope not! But maybe there should be (or are) functions for converting such values to _standard_ platform-independent bit representations... -- Ashley Yakeley, Seattle WA ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
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Re: Machine bit representations
[EMAIL PROTECTED] wrote: Hello, Is it possible in Haskell to access the underlying machine bit representation of a Float or Double value? I need to be able to be able to send this bit representation as a list of bytes in network byte order to a process running on a different platform (with a different host byte order to my platform). For reference, I run Haskell under Linux on Intel. The processes I want to communicate with run under Sun Solaris, Hitachi HPUX and Java everywhere. Any suggestions, for any Haskell translator much appreciated. Thanks. Look at encodeFloat and decodeFloat, documented in section 6.4.6 of The Haskell 98 Report. They offer efficient, machine-independent access to the components of a floating-point number. See http://www.haskell.org/onlinereport/basic.html#sect6.4.6 Matt Harden ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: Machine bit representations
On 2001-07-24T16:51:54-0700, [EMAIL PROTECTED] wrote: Hello, Is it possible in Haskell to access the underlying machine bit representation of a Float or Double value? I need to be able to be able to send this bit representation as a list of bytes in network byte order to a process running on a different platform (with a different host byte order to my platform). For reference, I run Haskell under Linux on Intel. The processes I want to communicate with run under Sun Solaris, Hitachi HPUX and Java everywhere. Any suggestions, for any Haskell translator much appreciated. If you're using GHC, you should be able to get at the raw bits of the machine representation using some Storable + Ptr + MarshallAlloc trickery. (These are modules in -package lang.) I tested the following on GHC5 on i386-linux: module Cast (cast) where import Storable (Storable, sizeOf, peek) import MarshalUtils (withObject) import IOExts (unsafePerformIO) import Ptr (castPtr) cast :: (Storable a, Storable b) = a - b cast a = b where b | sizeOf a == sizeOf b = unsafePerformIO $ withObject a $ peek . castPtr I was able to get (cast :: Int - Char) 98 == 'b' This might work with other compilers as well, but I don't know... Fun fun fun! More efficient implementations? -- Edit this signature at http://www.digitas.harvard.edu/cgi-bin/ken/sig « ne gâche pas ta vie pour leur idée de patrie » le bouton me dit PGP signature
