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Re: free variable
by Jon Fairbairn 09 Aug '01

09 Aug '01

> Hi, Hi. This sort of question belongs on haskell-café. I've copied this reply there. > I still have confusing regarding the following case: > > elemNum2 :: Int -> [Int] -> Int > elemNum2 elem list = sum [ 1 | findElem <- list, elem == findElem ] > > the function I wrote finds the number of occurences of an element in a list. > If I change it to the following: > > elemNum2 elem list = sum [ 1 | elem <- list ] > > Now it simply gives the length of the list. OK I understand that "elem" in > the list comprehension defines a new variable on the fly. What I do not > understand is in the first case: > > "findElem <- list" findElem is a new variable, but "list" is not. > "elem==findElem" here for some reason "elem" is not a new variable. > > Why does the rule only apply for "<-" operation, but not "==" for example? > > Thanks <- and let ... = ... aren't operations, they're definitions. == is the usual comparison operator. A list comprehension includes both definitions and predicates -- expressions of type Bool, or if you prefer, "operations returning" Bool. Definitions are either "let <var> = <expr>", as in [a | let a = 1], which evaluates to [1], or <var> <- <expr> where expr:: [t] for some t. You can read <- as "drawn from" [a | a <- [1,2,3]] evaluates to [1,2,3], and if we add a predicate: [a | a <- [1,2,3], odd a] evaluates to [1,3] and [a | a <- [1,2,3], a == 1] evaluates to [1] Hope that helps Jón -- Jón Fairbairn Jon.Fairbairn(a)cl.cam.ac.uk 31 Chalmers Road jf(a)cl.cam.ac.uk Cambridge CB1 3SZ +44 1223 570179 (after 14:00 only, please!)

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Re: Currying - Curry function
by Ashley Yakeley 08 Aug '01

08 Aug '01

At 2001-08-07 14:20, Mark Carroll wrote: >> Is there an exemple of such a function in the Haskell language ? > >Most certainly. For instance, we can define: > > f :: Integer -> Integer -> Integer > f x y = x + y > > g = f 1 That's not really currying, is it? This is currying: f :: (Integer,Integer) -> Integer f (x,y) = x + y g :: Integer -> Integer g y = f (1,y) -- g is f with the first argument curried with 1. h :: Integer -> Integer h x = f (x,1) -- h is f with the second argument curried with 1. And these are the functions for the first of two arguments: curry :: ((a,b) -> c) -> (a -> b -> c) curry f a b = f (a,b) uncurry :: (a -> b -> c) -> ((a,b) -> c) uncurry f (a,b) = f a b ... so that we can define g = curry f 1 -- Ashley Yakeley, Seattle WA

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Currying - Curry function
by Normand Leclerc 08 Aug '01

08 Aug '01

Hi haskell-cafe, >From a non programmer : Does someone know the exact source, book, page no etc. where Haskell Curry intoduced and developed its concept of a curry function. Is there an exemple of such a function in the Haskell language ? Thanks, Normand mailto:leclercn@globetrotter.net

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RE: Currying - Curry function
by Konst Sushenko 08 Aug '01

08 Aug '01

to check whether I myself understand currying correctly: currying is not partial application (as the example below seems to suggest), it is conversion of a function in to a form where partial application is possible. so we can say that g = curry f is a curried function. right? konst : -----Original Message----- : From: Ashley Yakeley [mailto:ashley@semantic.org] : Sent: Tuesday, August 07, 2001 3:46 PM : To: Mark Carroll; Haskell Cafe List : Subject: Re: Currying - Curry function : : : At 2001-08-07 14:20, Mark Carroll wrote: : : >> Is there an exemple of such a function in the Haskell language ? : > : >Most certainly. For instance, we can define: : > : > f :: Integer -> Integer -> Integer : > f x y = x + y : > : > g = f 1 : : That's not really currying, is it? This is currying: : : f :: (Integer,Integer) -> Integer : f (x,y) = x + y : : g :: Integer -> Integer : g y = f (1,y) : : -- g is f with the first argument curried with 1. : : h :: Integer -> Integer : h x = f (x,1) : : -- h is f with the second argument curried with 1. : : And these are the functions for the first of two arguments: : : curry :: ((a,b) -> c) -> (a -> b -> c) : curry f a b = f (a,b) : : uncurry :: (a -> b -> c) -> ((a,b) -> c) : uncurry f (a,b) = f a b : : ... so that we can define : g = curry f 1 : : : -- : Ashley Yakeley, Seattle WA : : : _______________________________________________ : Haskell-Cafe mailing list : Haskell-Cafe(a)haskell.org : http://www.haskell.org/mailman/listinfo/haskell-cafe :

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Missing functions in GHC's RTS?
by Ashley Yakeley 04 Aug '01

04 Aug '01

At 2001-08-03 00:41, I wrote (on the Haskell Cafe list): >When I make a function generator with "foreign export dynamic", GHC spits >out stub files which it apparently then needs for the link. What are >these for? OK, so I don't know what they are, but the link needs the _stub.o file. Unfortunately, the stub file includes these symbols: rts_getWord8 rts_getWord16 rts_getInt16 rts_getInt64 ...which do not appear to be in any library and indeed are not declared in RtsAPI.h, although corresponding rts_mk* functions are. This looks like a bug in GHC. $ ghc -v Glasgow Haskell Compiler, Version 5.00, for Haskell 98, compiled by GHC version 5.00 Using package config file: /usr/lib/ghc-5.00/package.conf Hsc static flags: -static -fignore-interface-pragmas -fomit-interface-pragmas -fdo-lambda-eta-expansion -flet-no-escape $ uname -a Linux server 2.2.19pre17 #1 Tue Mar 13 22:37:59 EST 2001 i686 unknown -- Ashley Yakeley, Seattle WA

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"foreign export dynamic" in GHC
by Ashley Yakeley 03 Aug '01

03 Aug '01

When I make a function generator with "foreign export dynamic", GHC spits out stub files which it apparently then needs for the link. What are these for? For instance: -- foreign export dynamic mkCallback :: (Int -> IO Int) -> IO Addr foreign import registerCallback :: Addr -> IO () exportCallback :: (Int -> IO Int) -> IO () exportCallback f = do fx <- mkCallback f registerCallback fx -- And then I expect the C side of things to look like this: -- typedef HsInt (*mkCallback_FunPtr) (HsInt arg1); void registerCallback (HsAddr f) { mkCallback_FunPtr callBack = reinterpret_cast<mkCallback_FunPtr>(f); ... // actually call the callback HsInt arg = 3; HsInt result = (*callBack)(arg); ... } -- Is this correct? When I compile the Haskell Test.hs I get these strange stub files, "Test_stub.h", "Test_stub.c" and "Test_stub.o". The _stub.h one looks like this: -- #include "HsFFI.h" extern HsInt Test_dvD(HsStablePtr a0, HsAddr a_, HsInt a1); -- ...while the _stub.c looks like this: -- #include "Stg.h" #include "HsStd.h" #include "HsLang.h" #include "RtsAPI.h" extern HaskellObj Test_zdfmkCallback_closure; HsInt Test_dvD(HsStablePtr a0, HsAddr a_, HsInt a1) { SchedulerStatus rc; HaskellObj ret; rc=rts_evalIO(rts_apply(rts_apply((StgClosure*)&Test_zdfmkCallback_closure, rts_mkStablePtr(a0)),rts_mkInt(a1)),&ret); rts_checkSchedStatus("Test_dvD",rc); return(rts_getInt(ret)); } -- What are the stubs for? The GHC manual, sec. 8.4.1 says this: # An FFI implementation is encouraged to generate the C # typedef corresponding to a foreign export dynamic # declaration, but isn't required to do so. It doesn't say anything about funny little stub functions. And curiously, it doesn't say exactly how GHC's FFI implementation responds to that non-requirement encouragement. There were no typedefs in the stub files. -- Ashley Yakeley, Seattle WA

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Another query (-:
by Mark Carroll 02 Aug '01

02 Aug '01

I've made a bizarre little program that I don't understand the behaviour of. It's: data BoolList = End | Node Bool BoolList instance Show BoolList where show (Node False rest) = "F " ++ show rest show (Node True rest) = "T " ++ show rest show End = "<>" grow list = (Node False rest) where (Node _ rest) = add list add End = (Node True End) add (Node truth rest) = (Node truth (add rest)) The show is fine; it's the add that's confusing me. Where we have, grow list = (Node False rest) where (Node _ rest) = add list I don't really understand what's going on with this binding and matching using "where", so I'm not seeing how the True and False values are managing to propagate up to being results of grow as they do, e.g.: > grow End F <> > grow (grow End) F T <> I have Hudak, Peterson, Fasel's gentle introduction to Haskell 98, and Thompson's "Craft of Functional Programming" book - maybe I just missed something in them? Is it easy to explain what's going on with this "where" behaviour so that I could figure out what "grow"'s results are without evaluating examples? -- Mark

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Re: Novice Haskell query
by Ashley Yakeley 02 Aug '01

02 Aug '01

At 2001-07-31 14:17, Mark Carroll wrote: >Now, "show x" will work, and "show y" won't. How can I write things so >that I can show both x and y? I don't seem to be able to write, say, > > instance Show a => Show (Tree a) where > show (Branch datum left right) = > show left ++ " <- " ++ show datum ++ " -> " ++ show right > > instance Show (Tree a) where > show Leaf = "*" These two instances overlap. You want one instance, with both Branch and Leaf alternatives in the 'show' function. >or, perhaps, > > instance Show (Tree a) where > Show a => show (Branch datum left right) = > show left ++ " <- " ++ show datum ++ " -> " ++ show right > show Leaf = "*" Close, but you're making a promise (that 'Tree a' is Show for _any_ 'a') that you can't fulfill. Attempting to restrict the type of 'a' while defining the function is bogus, the 'a' symbol simply isn't in scope at that point. Clearly the type 'Tree a' is only Show if 'a' is. It's irrelevant that the 'Leaf' part of the definition doesn't need 'a' to be Show: the 'Branch' part does, so the type has to be restricted to 'Show a => Tree a' if it is going to be Show. So you want this: instance Show a => Show (Tree a) where ... -- Ashley Yakeley, Seattle WA

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Re: newbie conceptual question [from haskell list]
by Steinitz, Dominic J 01 Aug '01

01 Aug '01

Alex, The code isn't ready to go public but you can try out a web interface at http://r341-02.king.ac.uk/cgi-bin/ldap.cgi. If you use the search filter objectClass=top you will get the whole tree on our test server. You shouldn't be able to add, modify or delete (let me know if you can!). The web code isn't very resilient. If you specifiy a non-existent attribute or use the wild card character (*), you will get an uninformative error message. Or you can try searching any public ldap server. Dominic. alex(a)shop.com on 01/08/2001 13:30:00 To: Dominic Steinitz cc: hdaume franka haskell-cafe bcc: Subject: Re: newbie conceptual question [from haskell list] Is the LDAP client available somewhere? -Alex- On 1 Aug 2001, Steinitz, Dominic J wrote: > I don't know about functional dependencies but using an existential type turned out to be very useful in writing an LDAP protocol handler. The protocol is specified at an abstract level using ASN.1 and could, in theory, be encoded using any set of encoding rules. It happens to use the Basic Encoding Rules. We used an existential type to "encode" the protocol at an abstract level and the encoding rules take this type and produce a concrete representation ready to send over a transport mechanism. Thus we get a good separation between the abstract protocol and the concrete encoding. So the next time we implement a protocol handler we can re-use the encoding code or we could encode LDAP with a different set of encoding rules without having to touch the LDAP code itself. > > We are presenting a paper which includes this at the Scottish Functional Programming workshop. > > Dominic. > > > > > hdaume(a)ISI.EDU on 31/07/2001 22:29:00 > To: franka > cc: haskell-cafe > bcc: Dominic Steinitz > Subject: Re: newbie conceptual question [from haskell list] > > Hi, > > Frank Atanassow wrote: > > D. Tweed wrote: > > > I've never written a Haskell program using functional dependencies, or > > > existential classes, ... > > > > I find them indispensible, and I know for a fact that I am not the only one > > around our office who feels that way. Though, the people around here > > (Utrecht's software technology group) are not exactly typical programmers... > > :) > > I've been recently experimenting quite a bit with existential classes > and somewhat less with functional dependencies, primarily to help my > understanding of the concepts. However, I've yet to be able to think of > an appropriate place to use them in the "real world". That is, in > something more than a toy thought-experiment. Could you give some > examples of what you are using them for? > > -- > Hal Daume III > > "Computer science is no more about computers | hdaume(a)isi.edu > than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > > > ------------------------------------------------------------------------------- ------------------ > 21st century air travel http://www.britishairways.com > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > ___________________________________________________________________ S. Alexander Jacobson Shop.Com 1-646-638-2300 voice The Easiest Way To Shop (sm) ------------------------------------------------------------------------------------------------- 21st century air travel http://www.britishairways.com

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Re: newbie conceptual question [from haskell list]
by S. Alexander Jacobson 01 Aug '01

01 Aug '01

Is the LDAP client available somewhere? -Alex- On 1 Aug 2001, Steinitz, Dominic J wrote: > I don't know about functional dependencies but using an existential type turned out to be very useful in writing an LDAP protocol handler. The protocol is specified at an abstract level using ASN.1 and could, in theory, be encoded using any set of encoding rules. It happens to use the Basic Encoding Rules. We used an existential type to "encode" the protocol at an abstract level and the encoding rules take this type and produce a concrete representation ready to send over a transport mechanism. Thus we get a good separation between the abstract protocol and the concrete encoding. So the next time we implement a protocol handler we can re-use the encoding code or we could encode LDAP with a different set of encoding rules without having to touch the LDAP code itself. > > We are presenting a paper which includes this at the Scottish Functional Programming workshop. > > Dominic. > > > > > hdaume(a)ISI.EDU on 31/07/2001 22:29:00 > To: franka > cc: haskell-cafe > bcc: Dominic Steinitz > Subject: Re: newbie conceptual question [from haskell list] > > Hi, > > Frank Atanassow wrote: > > D. Tweed wrote: > > > I've never written a Haskell program using functional dependencies, or > > > existential classes, ... > > > > I find them indispensible, and I know for a fact that I am not the only one > > around our office who feels that way. Though, the people around here > > (Utrecht's software technology group) are not exactly typical programmers... > > :) > > I've been recently experimenting quite a bit with existential classes > and somewhat less with functional dependencies, primarily to help my > understanding of the concepts. However, I've yet to be able to think of > an appropriate place to use them in the "real world". That is, in > something more than a toy thought-experiment. Could you give some > examples of what you are using them for? > > -- > Hal Daume III > > "Computer science is no more about computers | hdaume(a)isi.edu > than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > > > ------------------------------------------------------------------------------------------------- > 21st century air travel http://www.britishairways.com > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > ___________________________________________________________________ S. Alexander Jacobson Shop.Com 1-646-638-2300 voice The Easiest Way To Shop (sm)

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