December 2010 - Beginners

dealing with chained functions
by Alec Benzer 20 Apr '11

20 Apr '11

Is there a generally accepted "better" way of dealing with the following: say I have three functions that all rely on output from each other. Ie, we get raw input, which one function processes. another function then requires the output from this function and produces some other output, which a third function then requires. would it be better to implement the functions like this: f1 :: a -> b f2 :: b -> c f3 :: c -> d final :: a -> d or like this: f1 :: a -> b f2 :: a -> c f3 :: a -> d Ie, in the latter set, f2 calls f1 on its input, and then proceeds to process it. f3 calls f2 on its input, etc. In the former case, final would just chain all three functions together itself (or I might just do it in main or somewhere else that's appropriate) Are one of these methods generally considered a better practice than the other, specifically in a functional context? My gut tells me that the former method looks more appropriate, but I'm not sure if it'd really be the accepted practice regardless of context, or if one might use either of these methods depending on the specifics of what they're doing. Or if it's just a matter of taste/opinion?

4 6

Tying the knot
by Alex Rozenshteyn 05 Jan '11

05 Jan '11

I'm trying to understand the technique referred to as "tying the knot", but documentation on the internet seems to be much sparser and more obtuse than I like. So I'm asking here. As far as I understand, "tying the knot" refers to a way of using laziness to implement something like references in a purely functional way. I'm trying to write a toy simulation: I have a population :: [Person] I want to collect a random subset of possible pairs of distinct people. So I go to each person in the population and select a subset of the people after him/her in the list; these are pairs in which s/he is the first element. I want to then be able to ask for all pairs in which a person is the first or the second element. I could give each person a unique id, but it seems like tying the knot is a valid way to implement this. Please correct me if I am misunderstanding. Thank you. -- Alex R

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Re: [Haskell-beginners] Beginners Digest, Vol 30, Issue 49
by Miguel Dias 31 Dec '10

31 Dec '10

i have the version 6.10.4. i realized now, that has a new version, i've tried to install it, but it still's in the old version. maybe i need to install cabal. i've downloaded of my actual version from the www.haskell.org/ghc Miguel On Dec 31, 2010, at 11:00 AM, beginners-request(a)haskell.org wrote: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://www.haskell.org/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. Problem installing lhs2tex (Miguel Dias) > 2. Re: Problem installing lhs2tex (Brent Yorgey) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Thu, 30 Dec 2010 19:29:38 +0000 > From: Miguel Dias <dias.69.652(a)gmail.com> > Subject: [Haskell-beginners] Problem installing lhs2tex > To: beginners(a)haskell.org > Message-ID: > <AANLkTimbk3ePJ2Sn-ed472bqGaTAnUGyv8abqJAK7v3p(a)mail.gmail.com> > Content-Type: text/plain; charset="iso-8859-1" > > Hi, im trying to install lhs2TeX in Mac OS. > But i got an error, that i dont know how to fix. > > Someone can explain me what i need to do to fix that. > > Is this error: > > " > MacBook:lhs2tex-1.16 mgd$ make > /opt/local/bin/ghc -O --make -o lhs2TeX Main.lhs > [17 of 19] Compiling FileNameUtils ( FileNameUtils.lhs, FileNameUtils.o ) > > FileNameUtils.lhs:15:57: > Module `System.IO' does not export `hSetEncoding' > > FileNameUtils.lhs:15:85: Module `System.IO' does not export `utf8' > make: *** [lhs2TeX] Error 1 > > " > > Thanks, > Miguel >

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Problem installing lhs2tex
by Miguel Dias 30 Dec '10

30 Dec '10

Hi, im trying to install lhs2TeX in Mac OS. But i got an error, that i dont know how to fix. Someone can explain me what i need to do to fix that. Is this error: " MacBook:lhs2tex-1.16 mgd$ make /opt/local/bin/ghc -O --make -o lhs2TeX Main.lhs [17 of 19] Compiling FileNameUtils ( FileNameUtils.lhs, FileNameUtils.o ) FileNameUtils.lhs:15:57: Module `System.IO' does not export `hSetEncoding' FileNameUtils.lhs:15:85: Module `System.IO' does not export `utf8' make: *** [lhs2TeX] Error 1 " Thanks, Miguel

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Re: [Haskell-beginners] Tying the knot
by Russ Abbott 29 Dec '10

29 Dec '10

I haven't done any prolog in a while, but I thought this reminded me of prolog variables. So I wrote up a prolog version. This is the code. labelLeaves(Tree, Tree1) :- label(N, Tree, (N, Tree1)). label(N, branch(A, B), (N1, branch(A1, B1))) :- label(N, A, (Na, A1)), label(N, B, (Nb, B1)), N1 is Na + Nb. label(N, leaf(_), (1, leaf(N))). # The next line provides data to work with. tree(branch(branch(leaf(a), branch(leaf(b), leaf(c))), leaf(d))). Here is the execution. ?- tree(T), labelLeaves(T, T1). T = branch(branch(leaf(a), branch(leaf(b), leaf(c))), leaf(d)), T1 = branch(branch(leaf(4), branch(leaf(4), leaf(4))), leaf(4)) ; false. * -- Russ Abbott _____________________________________________* * Professor, Computer Science California State University, Los Angeles Google voice: 424-235-5752 (424-cell-rja) blog: http://russabbott.blogspot.com/ vita: http://sites.google.com/site/russabbott/ _____________________________________________* On Wed, Dec 29, 2010 at 4:22 PM, <beginners-request(a)haskell.org> wrote: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://www.haskell.org/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. Re: strange behaviour : computing lowest divisor (Daniel Fischer) > 2. Re: Tying the knot (Heinrich Apfelmus) > 3. runtime error <<loop>> when using -O compile option > (Gerold Meisinger) > 4. Re: runtime error <<loop>> when using -O compile option > (Sangeet Kumar) > 5. Re: Tying the knot (Alex Rozenshteyn) > 6. Re: Tying the knot (Patrick LeBoutillier) > 7. Re: Tying the knot (aditya siram) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Wed, 29 Dec 2010 12:49:00 +0100 > From: Daniel Fischer <daniel.is.fischer(a)googlemail.com> > Subject: Re: [Haskell-beginners] strange behaviour : computing lowest > divisor > To: beginners(a)haskell.org > Message-ID: <201012291249.01415.daniel.is.fischer(a)googlemail.com> > Content-Type: text/plain; charset="utf-8" > > On Wednesday 29 December 2010 10:59:53, Abhijit Ray wrote: > > Thanks, that seems to have fixed it. > > > > On Wed, Dec 29, 2010 at 5:46 PM, Lyndon Maydwell <maydwell(a)gmail.com> > wrote: > > > Try with Integer rather than Int. Might be an overflow issue... > > > > > > > > > > > > *Main> ld 278970415063349480483707695 > > Yes, that number is between 2^87 and 2^88, as an Int, it's typically one of > 206297903 = 7*37*796517 (32-bit Ints) > or > -9158009321667437777 = -7*13*15015953*670203649 (64-bit Ints). > > > > ------------------------------ > > Message: 2 > Date: Wed, 29 Dec 2010 13:49:34 +0100 > From: Heinrich Apfelmus <apfelmus(a)quantentunnel.de> > Subject: Re: [Haskell-beginners] Tying the knot > To: beginners(a)haskell.org > Message-ID: <iffaou$ba4$1(a)dough.gmane.org> > Content-Type: text/plain; charset=UTF-8; format=flowed > > Alex Rozenshteyn wrote: > > I'm trying to understand the technique referred to as "tying the knot", > but > > documentation on the internet seems to be much sparser and more obtuse > than > > I like. > > > > So I'm asking here. > > > > As far as I understand, "tying the knot" refers to a way of using > laziness > > to implement something like references in a purely functional way. > > Not really. > > "Tying the knot" refers to writing a seemingly circular program, where > the result of a function is used as argument to the very same function. > > A canonical example is the following solution to the problem of labeling > all the leaves in a tree with the total leaf count: > > data Tree a = Branch (Tree a) (Tree a) | Leaf a > > labelLeaves :: Tree a -> Tree Int > labelLeaves tree = tree' > where > (n, tree') = label n tree -- n is both result and argument! > > label n (Branch a b) = (na+nb, Branch a' b') > where > (na,a') = label n a > (nb,b') = label n b > label n (Leaf _) = (1, Leaf n) > > > In some cases, this be used to implement read-only doubly-linked lists > and other things that seem to require references, but not everything > involving references can be solved by tying a knot; in particular, the > references will be read-only. > > > (Feel free to put my blurb above on the HaskellWiki.) > > > I'm trying to write a toy simulation: > > I have a population :: [Person] > > I want to collect a random subset of possible pairs of distinct people. > > So I go to each person in the population and select a subset of the > people > > after him/her in the list; these are pairs in which s/he is the first > > element. > > > > I want to then be able to ask for all pairs in which a person is the > first > > or the second element. I could give each person a unique id, but it > seems > > like tying the knot is a valid way to implement this. > > This situation doesn't have anything to do with tying the knot. After > all, how do you distinguish persons in the first place? You probably > already gave the unique IDs. Then, it's simply a matter of applying the > function > > filter (\(x,y) -> x == p || y == p) > > where p is the person you are looking for. > > > Regards, > Heinrich Apfelmus > > -- > http://apfelmus.nfshost.com > > > > > ------------------------------ > > Message: 3 > Date: Wed, 29 Dec 2010 13:57:15 +0000 (UTC) > From: Gerold Meisinger <gerold.meisinger(a)gmail.com> > Subject: [Haskell-beginners] runtime error <<loop>> when using -O > compile option > To: beginners(a)haskell.org > Message-ID: <loom.20101229T145705-480(a)post.gmane.org> > Content-Type: text/plain; charset=us-ascii > > Hello! > > I'm working on a computer game using Yampa and I get the following > runtime error: > > $ myprog: <<loop>> > > when compiling with > > $ ghc --make MyProg.hs -o myprog -O > (without -O it works fine) > > I stripped the bug down to the program below. What's funny is that the > error disappears under certain "odd circumstances" (marked as #1-#4). My > questions are: > 1. How can I avoid this bug without introducing one of the "odd > circumstances"? > 2. Why is it that I get this error? > 3. How would you hunt down such a bug? Originally I got no clue where it > came from, so I just took the program apart piece by piece. > > {-# LANGUAGE Arrows #-} > > module Main (main) where > > import FRP.Yampa > > type ObjIn = Event () -- loop #1 > --type ObjIn = Bool -- no loop #1 > > type ObjOut = (String, Int) -- loop #2 > --type ObjOut = Int -- no loop #2 > > type GameObj = SF ObjIn ObjOut > > testObj :: GameObj > testObj = proc hit -> do > returnA -< ("testObj", 1) -- loop #2 > -- returnA -< 1 -- no loop #2 > > process :: [GameObj] -> SF () [ObjOut] > process objs = proc _ -> do > rec > gamestate <- par logic objs > -< gamestate -- loop #3 (recursive definition!) > -- -< [] -- no loop #3 > > returnA -< gamestate > > logic :: [ObjOut] -> [sf] -> [(ObjIn, sf)] > logic gamestate objs = map route objs > where > route obj = > (if null (foo gamestate) then NoEvent else NoEvent, obj) > -- loop #1 > -- (if null (foo gamestate) then False else False, obj) > -- no loop #1 > > foo :: [ObjOut] -> [ObjOut] > foo [] = [] > foo objs = concat (collisions objs) > where > collisions [] = [] > collisions (out:objs') = > [[out, out'] | out' <- objs, out `collide` out'] -- loop > #4 > -- [[out, out'] | out' <- objs, True] -- no loop #4 > > collide :: ObjOut -> ObjOut -> Bool > collide (_, p) (_, p') = True -- loop #2 > --collide p p' = True -- no loop #2 > > > main :: IO () > main = do > putStrLn . show $ embed (process [testObj]) ((), [(1.0, > Nothing)]) > > (Btw: I re-opened a bug report: > http://hackage.haskell.org/trac/ghc/ticket/2722#comment:10 ) > > > > > > ------------------------------ > > Message: 4 > Date: Wed, 29 Dec 2010 15:06:41 +0100 (CET) > From: Sangeet Kumar <sk(a)one.com> > Subject: Re: [Haskell-beginners] runtime error <<loop>> when using -O > compile option > To: Gerold Meisinger <gerold.meisinger(a)gmail.com> > Cc: beginners(a)haskell.org > Message-ID: <10654318.18.1293631600822.JavaMail.sangeetk@sk> > Content-Type: text/plain; charset=utf-8 > > Hi, > > Thankyou for the update. I will confirm the delivery as soon as I receive > it. > > Regards, > Sangeet > > ----- Original Message ----- > From: "Gerold Meisinger" <gerold.meisinger(a)gmail.com> > To: beginners(a)haskell.org > Sent: Wednesday, December 29, 2010 2:57:15 PM > Subject: [Haskell-beginners] runtime error <<loop>> when using -O compile > option > > Hello! > > I'm working on a computer game using Yampa and I get the following > runtime error: > > $ myprog: <<loop>> > > when compiling with > > $ ghc --make MyProg.hs -o myprog -O > (without -O it works fine) > > I stripped the bug down to the program below. What's funny is that the > error disappears under certain "odd circumstances" (marked as #1-#4). My > questions are: > 1. How can I avoid this bug without introducing one of the "odd > circumstances"? > 2. Why is it that I get this error? > 3. How would you hunt down such a bug? Originally I got no clue where it > came from, so I just took the program apart piece by piece. > > {-# LANGUAGE Arrows #-} > > module Main (main) where > > import FRP.Yampa > > type ObjIn = Event () -- loop #1 > --type ObjIn = Bool -- no loop #1 > > type ObjOut = (String, Int) -- loop #2 > --type ObjOut = Int -- no loop #2 > > type GameObj = SF ObjIn ObjOut > > testObj :: GameObj > testObj = proc hit -> do > returnA -< ("testObj", 1) -- loop #2 > -- returnA -< 1 -- no loop #2 > > process :: [GameObj] -> SF () [ObjOut] > process objs = proc _ -> do > rec > gamestate <- par logic objs > -< gamestate -- loop #3 (recursive definition!) > -- -< [] -- no loop #3 > > returnA -< gamestate > > logic :: [ObjOut] -> [sf] -> [(ObjIn, sf)] > logic gamestate objs = map route objs > where > route obj = > (if null (foo gamestate) then NoEvent else NoEvent, obj) > -- loop #1 > -- (if null (foo gamestate) then False else False, obj) > -- no loop #1 > > foo :: [ObjOut] -> [ObjOut] > foo [] = [] > foo objs = concat (collisions objs) > where > collisions [] = [] > collisions (out:objs') = > [[out, out'] | out' <- objs, out `collide` out'] -- loop > #4 > -- [[out, out'] | out' <- objs, True] -- no loop #4 > > collide :: ObjOut -> ObjOut -> Bool > collide (_, p) (_, p') = True -- loop #2 > --collide p p' = True -- no loop #2 > > > main :: IO () > main = do > putStrLn . show $ embed (process [testObj]) ((), [(1.0, > Nothing)]) > > (Btw: I re-opened a bug report: > http://hackage.haskell.org/trac/ghc/ticket/2722#comment:10 ) > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > > > ------------------------------ > > Message: 5 > Date: Wed, 29 Dec 2010 14:31:48 -0500 > From: Alex Rozenshteyn <rpglover64(a)gmail.com> > Subject: Re: [Haskell-beginners] Tying the knot > To: Heinrich Apfelmus <apfelmus(a)quantentunnel.de> > Cc: beginners(a)haskell.org > Message-ID: > <AANLkTi=T4QYGbx9r_xoD8e8cPiOak58DvMqZZNta=wHL(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > Thank you. I'm still unclear as to how tying the know works and when it is > useful, but at least one of my misconceptions has been clarified. > > I haven't given my `Person`s unique ids yet, thinking that I could avoid it > if I worked carefully. Guess it's not worth the effort. > > On Wed, Dec 29, 2010 at 7:49 AM, Heinrich Apfelmus < > apfelmus(a)quantentunnel.de> wrote: > > > Alex Rozenshteyn wrote: > > > >> I'm trying to understand the technique referred to as "tying the knot", > >> but > >> documentation on the internet seems to be much sparser and more obtuse > >> than > >> I like. > >> > >> So I'm asking here. > >> > >> As far as I understand, "tying the knot" refers to a way of using > laziness > >> to implement something like references in a purely functional way. > >> > > > > Not really. > > > > "Tying the knot" refers to writing a seemingly circular program, where > the > > result of a function is used as argument to the very same function. > > > > A canonical example is the following solution to the problem of labeling > > all the leaves in a tree with the total leaf count: > > > > data Tree a = Branch (Tree a) (Tree a) | Leaf a > > > > labelLeaves :: Tree a -> Tree Int > > labelLeaves tree = tree' > > where > > (n, tree') = label n tree -- n is both result and argument! > > > > label n (Branch a b) = (na+nb, Branch a' b') > > where > > (na,a') = label n a > > (nb,b') = label n b > > label n (Leaf _) = (1, Leaf n) > > > > > > In some cases, this be used to implement read-only doubly-linked lists > and > > other things that seem to require references, but not everything > involving > > references can be solved by tying a knot; in particular, the references > will > > be read-only. > > > > > > (Feel free to put my blurb above on the HaskellWiki.) > > > > > > I'm trying to write a toy simulation: > >> I have a population :: [Person] > >> I want to collect a random subset of possible pairs of distinct people. > >> So I go to each person in the population and select a subset of the > people > >> after him/her in the list; these are pairs in which s/he is the first > >> element. > >> > >> I want to then be able to ask for all pairs in which a person is the > first > >> or the second element. I could give each person a unique id, but it > seems > >> like tying the knot is a valid way to implement this. > >> > > > > This situation doesn't have anything to do with tying the knot. After > all, > > how do you distinguish persons in the first place? You probably already > gave > > the unique IDs. Then, it's simply a matter of applying the function > > > > filter (\(x,y) -> x == p || y == p) > > > > where p is the person you are looking for. > > > > > > Regards, > > Heinrich Apfelmus > > > > -- > > http://apfelmus.nfshost.com > > > > > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org > > http://www.haskell.org/mailman/listinfo/beginners > > > > > > -- > Alex R >

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runtime error <<loop>> when using -O compile option
by Gerold Meisinger 29 Dec '10

29 Dec '10

Hello! I'm working on a computer game using Yampa and I get the following runtime error: $ myprog: <<loop>> when compiling with $ ghc --make MyProg.hs -o myprog -O (without -O it works fine) I stripped the bug down to the program below. What's funny is that the error disappears under certain "odd circumstances" (marked as #1-#4). My questions are: 1. How can I avoid this bug without introducing one of the "odd circumstances"? 2. Why is it that I get this error? 3. How would you hunt down such a bug? Originally I got no clue where it came from, so I just took the program apart piece by piece. {-# LANGUAGE Arrows #-} module Main (main) where import FRP.Yampa type ObjIn = Event () -- loop #1 --type ObjIn = Bool -- no loop #1 type ObjOut = (String, Int) -- loop #2 --type ObjOut = Int -- no loop #2 type GameObj = SF ObjIn ObjOut testObj :: GameObj testObj = proc hit -> do returnA -< ("testObj", 1) -- loop #2 -- returnA -< 1 -- no loop #2 process :: [GameObj] -> SF () [ObjOut] process objs = proc _ -> do rec gamestate <- par logic objs -< gamestate -- loop #3 (recursive definition!) -- -< [] -- no loop #3 returnA -< gamestate logic :: [ObjOut] -> [sf] -> [(ObjIn, sf)] logic gamestate objs = map route objs where route obj = (if null (foo gamestate) then NoEvent else NoEvent, obj) -- loop #1 -- (if null (foo gamestate) then False else False, obj) -- no loop #1 foo :: [ObjOut] -> [ObjOut] foo [] = [] foo objs = concat (collisions objs) where collisions [] = [] collisions (out:objs') = [[out, out'] | out' <- objs, out `collide` out'] -- loop #4 -- [[out, out'] | out' <- objs, True] -- no loop #4 collide :: ObjOut -> ObjOut -> Bool collide (_, p) (_, p') = True -- loop #2 --collide p p' = True -- no loop #2 main :: IO () main = do putStrLn . show $ embed (process [testObj]) ((), [(1.0, Nothing)]) (Btw: I re-opened a bug report: http://hackage.haskell.org/trac/ghc/ticket/2722#comment:10 )

2 1

strange behaviour : computing lowest divisor
by Abhijit Ray 29 Dec '10

29 Dec '10

HI, I have written(copied) a small piece of code which finds the lowest divisor of am integer (greater than 1) the code is here ld::Int->Int ld n = ld' 2 n where ld' i n | rem n i == 0 = i | i^2 > n = n | otherwise = ld' (i+1) n now it seems to work fine for small numbers but for a big number we get *Main> ld 278970415063349480483707695 7 the number is obviously divisible by 5 , so where is the anomaly ? Thanks, Abhijit On Wed, Dec 29, 2010 at 2:52 PM, Alex Rozenshteyn <rpglover64(a)gmail.com>wrote: > I'm trying to understand the technique referred to as "tying the knot", but > documentation on the internet seems to be much sparser and more obtuse than > I like. > > So I'm asking here. > > As far as I understand, "tying the knot" refers to a way of using laziness > to implement something like references in a purely functional way. > > I'm trying to write a toy simulation: > I have a population :: [Person] > I want to collect a random subset of possible pairs of distinct people. > So I go to each person in the population and select a subset of the people > after him/her in the list; these are pairs in which s/he is the first > element. > > I want to then be able to ask for all pairs in which a person is the first > or the second element. I could give each person a unique id, but it seems > like tying the knot is a valid way to implement this. > > Please correct me if I am misunderstanding. > > Thank you. > > -- > Alex R > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > >

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lambda
by John Moore 28 Dec '10

28 Dec '10

Hi, Can someone explain how this lambda is suppose to work, I cannot figure out what it does? data Lambda = Val Double | Add Lambda Lambda | Subtract Lambda Lambda | Multiply Lambda Lambda | Divide Lambda Lambda | Var String | Let String Lambda Lambda deriving Show demo1 = (Add(Multiply(Divide(Subtract(Val 25)(Val 5))(Val 10))(Val 7))(Val 30)) --let x = 1+1 in 3 - x test1 = Let "x" (Add (Val 1) (Val 1)) (Add(Var "x")(Var "x")) test6 = Let "y" (Add (Val 7)(Val 6)) (Subtract (Val 6)(Var "y")) -- 4 * (let x = 1+1 in 3 + x) test2 = Multiply (Val 4) test1 -- x * (let x = 1+1 in 3 + x) test3 = Let "y" (Add (Val 4)(Val 7))(Multiply test1 (Var "y")) test5 = Add (test1) test6 type Dict =[(String,Lambda)]-- dictionary which takes a string and an expression emptyDict :: Dict emptyDict = []-- empty dictionary null set addEntry :: String->Lambda ->Dict -> Dict addEntry= \n e d -> (n,e): d -- add an entry we take a string,expression and a dictionary :gives back dictionary lookupEntry :: String -> Dict -> Maybe Lambda lookupEntry n [] = Nothing lookupEntry n (x:xs) = if (n == k)-- if the entry is equal to then (Just v) else lookupEntry n xs -- find the value of n in the rest of the list where (k,v) = x -- evalStep :: Dict -> Lambda -> Lambda evalStep d (Val x) = (Val x)-- we have a dictionary and a value return the value evalStep d(Add x y)-- dictionary and a add expression of x y = case x of (Val a) -> case y of (Val b) -> Val (a+b) left -> Add x (evalStep d y) right -> Add (evalStep d x)y evalStep d(Subtract x y) = case x of (Val a) -> case y of (Val b) -> Val (a-b) left -> Subtract x (evalStep d y) right -> Subtract (evalStep d x)y evalStep d(Multiply x y) = case x of (Val a) -> case y of (Val b) -> Val (a*b) left -> Multiply x (evalStep d y) right -> Multiply (evalStep d x)y evalStep d (Divide x y) = case x of (Val a) -> case y of (Val b) -> Val (a/b) left -> Divide x (evalStep d y) right -> Divide (evalStep d x)y evalStep d (Let n e1 e2) = case e1 of (Val a) -> case e2 of (Val b)-> Val b left -> evalStep (addEntry n e1 d)e2 right -> evalStep (addEntry n e1 d) e2 evalStep d (Var x) = case lookup x d of Just e -> e Nothing -> error "Error in expression -- no definition for variable!" evaluate :: Dict-> [Lambda] -> Lambda -> IO() evaluate d(x:xs) e = do putStrLn (show e) putStrLn "Do another step (y/n) or rollback (r)? :" c <- getLine case c of "y" -> let e'= (evalStep d e)in evaluate d (e:x:xs) e'-- build up history "r" -> case (x:xs) of (x:xs)-> evaluate d xs x []-> do { putStrLn "Empty" ;evaluate d(x:xs) e } "n" -> putStrLn $ "Ok you said no :" ++ c ev :: Dict-> [Lambda] -> Lambda -> IO() ev = \d (x:xs) e ->do putStrLn (show e) putStrLn "Do another step (y/n) or rollback (r)? :" c <- getLine case c of "y" -> let e'= (evalStep d e)in evaluate d (e:x:xs) e'-- build up history "r" -> case (x:xs) of (x:xs)-> evaluate d xs x []-> do { putStrLn "Empty" ;evaluate d(x:xs) e } "n" -> putStrLn $ "Ok you said no :" ++ c isFree :: String -> Lambda -> Bool isFree n1 (Let n2 e1 e2) |n1 == n2 = False |otherwise = True subst :: String -> Lambda -> Lambda->Lambda subst n e1 e2 |isFree n e2 = subst' n e1 e2 |otherwise = e2 subst' :: String -> Lambda -> Lambda->Lambda subst' n0 e1 (Var n1) |n0 == n1 = e1 |otherwise = Var n1 subst' s v (Add e1 e2) = Add(subst' s v e1)(subst' s v e2) subst' s v (Multiply e1 e2) = Multiply(subst' s v e1)(subst' s v e2) subst' s v (Divide e1 e2) = Divide(subst' s v e1)(subst' s v e2) subst' s v (Subtract e1 e2) = Subtract(subst' s v e1)(subst' s v e2) subst' s v (Let n e1 e2) = Let n (subst' s v e1)(subst' s v e2) John

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Infix operator and precedence
by Patrick LeBoutillier 24 Dec '10

24 Dec '10

Hi, I'm just messing around with infix operators, and I'm trying to implement some kind of ternary if operator: (??) :: Bool -> (Bool -> a) -> a (??) = flip ($) (?:) :: a -> a -> (Bool -> a) t ?: f = \b -> if b then t else f test b = b ?? ("IF" ?: "THEN") I would like to get rid of the parentheses around the ("IF" ?: "THEN") part, but I'm not familiar with the use of theinfixr/infixl functions. I'm not sure which one to use in this case and how exactly to use it. Can anyone shed some light? Thanks a lot, Patrick -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

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how to input a Let expression
by John Moore 22 Dec '10

22 Dec '10

Hi, I'm missing something basic here, I wrote this program a while back. But I cant remember how to work it. Would someone please explain how I enter a value for a variable x,y etc and get this to work. It goes by entering evaluate [][test1]test1 but I cant remember how to write in The Let part and not have it keep ending in -no definotion of a variable. John

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