- Haskell-Cafe - Haskell.org

Newbie qustion about monads
by Juanma Barranquero 05 Oct '03

05 Oct '03

I have an extremely-newbie question about monads and how to interpret the monadic laws; I asked that same question yesterday on IRC and the answers were interesting but non-conclusive (to me anyway). I'm trying to learn monads by reading "All About Monads", version 1.0.2. I though of defining a monad like that: data Counted a = Ct Int a instance Monad (Counted) where return x = Ct 1 x Ct n x >>= f = let (Ct n' x') = f x in Ct (n+n') x' -- fail as default The intent is that Counted objects "count" the number of times an operation is applied to them. (For the purpose of the question, that's irrelevant anyway: the problem would be the same if I assigned a random number to the Int on each calling of >>=.) According to "All About Monads", the first monadic law says: (return x) >>= f == f x In my case, let's suppose I define: reverseCounted :: [a] -> Counted [a] reverseCounted x = return (reverse x) so I do: c1 = return "xxx" >>= reverseCounted -- c1 == Ct 2 "xxx" c2 = reverseCounted "xxx" -- c2 == Ct 1 "xxx" Now comes the question: In what sense should I interpret the "==" in the monadic law? Obviously, c1 and c2 are not structurally equal. If I can accept that two Counted things are "equal" even if they are not identical, is enough for me to define: instance Eq a => Eq (Counted a) where Ct _ x == Ct _ y = x == y to satisfy the first law? Yesterday I was said that it'd work if c1 and c2 are mutually substitutable. They are, for the purposes of equality, though evidently not for every purpose. A simple: count (Ct n _) = n allows me to distinguish between them. Juanma

5 13

Why does this work - haskell mysteries?!
by ,,, 04 Oct '03

04 Oct '03

Hi; the proof of the pudding does lies in the eating... but I still wonder why this code is working (it is taken from the book "The Craft of functional programming"). The program connects a variable-name to value. The fun initial gives the initial state, update sets a variable & value reads a value). I evaluate value my_store 'b' to 5 and value my_store 'a' to 3 as expected from the text in the book. But I can't see what is happening here. The book has a parallel example where the data is held in a list, and this version is easy to follow, but this trick with storing a lambda-function inside a newtype beats me. The problem is that I do not understand where the accumulated data is stored (not in a list - it seems like something like a chain of functions which can be pattern-matched, but I am not sure). And why does not the lambda-function (\w -> if v==w then n else sto w) start a endless loop? (This is not homework - I am a programmer who is curious about Haskell!) Any clues, anyone? Cheers, Petter -- Var is the type of variables. type Var = Char newtype Store = Sto (Var -> Int) -- initial :: Store initial = Sto (\v -> 0) value :: Store -> Var -> Int value (Sto sto) v = sto v update :: Store -> Var -> Int -> Store update (Sto sto) v n = Sto (\w -> if v==w then n else sto w) -- testit -- my_store = update (update (update initial 'a' 4) 'b' 5) 'a' 3)

1 0

Haskellsupport in KDevelop
by Peter Robinson 04 Oct '03

04 Oct '03

Hello, I've begun to write a plugin that provides basic support for Haskell in KDevelop 3.0 alpha. (http://www.kdevelop.org) It is already included in the CVS and the latest alpha7 release. Screenshots: http://www.thaldyron.com/snap1.png http://www.thaldyron.com/snap2.png http://www.thaldyron.com/snap3.png At the moment it's only possible to create a new project and to build/run it with GHC. What's really missing is a (primitive) background parser written that reports syntax errors. It can be written in yacc, antlr, etc., anything that produces C/C++ code. The only parsers for Haskell I could find are written themselves in Haskell. Does anyone know about one or must I write it from scratch? Regards, Peter

3 3

RE: "interact" behaves oddly if used interactively
by Simon Marlow 03 Oct '03

03 Oct '03

> Yes, the presence of lazy IO makes optimistic evaluation more > complicated, but I do not see that it compromises the purity of the > language in anyway (whatever purity is ;-). So, we're agreed that the presence of lazy evaluation makes implementing optimistic evaluation (and other evaluation strategies) more complicated. Personally, I find it disturbing that the presence of this family of library functions affects something as low-level as the evaluation strategy of the language. Or, to put it another way, it enforces a particular evaluation strategy on a part of the language, when the rest of Haskell makes no such restrictions. I would expect, in a pure language, that I could reduce any expression in a program to HNF (or _|_) without affecting the meaning of the program. In Haskell, I can only do this so long as the expression doesn't involve any lazy I/O. The real problem is that lazy I/O injects side effects into the pure world of expressions. Haskell has a perfectly good system for encapsulating side effects - the IO monad. So why put these sneaky side effects into pure values? Cheers, Simon

5 4

RE: "interact" behaves oddly if used interactively
by Simon Marlow 02 Oct '03

02 Oct '03

> > The real problem is that lazy I/O injects side effects into the pure > > world of expressions. Haskell has a perfectly good system for > > encapsulating side effects - the IO monad. So why put > these sneaky side > > effects into pure values? > > I fear one problem is that the word "side effect" is not properly > defined. > > Actually both getContents and interact are in the IO monad. Since the > action "interact" is not terminated until the input stream ends, I do > not see any problem there. Yes, it is probably true that you cannot observe any non-referential-transparency using only interact. The side effects are only visible in the IO monad, and since as you say interact doesn't let you back into the IO monad until it has finished, it is probably safe. The unsafety is only visible using the other lazy I/O operations, principally hGetContents. > Evaluation never leaves the IO monad until > termination. getContents is more strange. However, you can just define > its semantics to yield a random string. When you are not > careful you may certainly get something nearly random ;-) So... you agree that getContents in its current form should really be called unsafeGetContents? Unless perhaps we redefine its semantics to either (a) yield a random string or (b) eagerly slurp the entire stream contents? Cheers, Simon

4 3

RE: "interact" behaves oddly if used interactively
by Simon Marlow 02 Oct '03

02 Oct '03

[ taking this one to haskell-café... ] > I still do not quite agree with Simon that 'interact' exposes anything > but non-strictness. Non-strictness means that > > map toUpper _|_ = _|_ > map toUpper ('a':_|_) = ('A':_|_) > map toUpper ('a':'b':_|_) = ('A':'B':_|_) > > and 'interact (map toUpper)' is a great way to experience > this property. > > However, you can also experience the property without 'interact', > evaluating expressions like > > take 2 (map toUpper ('a':'b':undefined)) Certainly you can observe non-strictness, that's not the point. The point is that you can also observe more than just non-strictness using interact, and I don't think that is desirable. For example: interact (\xs -> let z = length xs in "Hello World\n") Now, Haskell is a pure language, so it shouldn't matter whether the implementation evaluates z or not, as long as it is careful not to violate the non-strict semantics by turning a terminating program into a non-terminating one. A parallel Haskell implementation might happily spawn off another thread to evaluate z for example. An optimistic implementation might evaluate z for a fixed amount of time before continuing with the main thread of evaluation. BUT in the presence of lazy I/O this simply isn't true any more. Why? Because z is not pure; evaluating it has a side-effect. And yet it has type Int. Am I the only one who thinks this is wrong? Cheers, Simon

8 11

Question about Type Checking
by Clement 02 Oct '03

02 Oct '03

Hi i am doing a type checker. and how can i check if the Expn that i have is a Prim Function = ("f",[([Arg "x"],Simple (Int 2))]) type Function = (Name, [Equation]) type Equation = ([Expn], Body) {- (the body of) either a simple equation: f a1 .. an = e or a conditional equation: f a1 .. an | b1 = e1 ... | bm = em data Body = Simple Expn | Conditional [(Expn, Expn)] deriving (Show, Read) data Expn = Int Int | Char Char | Bool Bool | Arg Name | Prim Name | Func Name | Appn Expn Expn | Tuple [Expn] and how can i group the Expn together for Appn? __________________________________ Do you Yahoo!? The New Yahoo! Shopping - with improved product search http://shopping.yahoo.com

1 0

Pre-Call for Contributions -- HC&A Report (November 2003 edition)
by C.Reinke 02 Oct '03

02 Oct '03

[apologies for multiple copies - experience has shown that not all Haskellers can be reached via the main Haskell list anymore] An entry for your diaries (no other action required right now): ------------------------------Pre-Call-------------------------------- Your contributions to the November 2003 edition of the Haskell Communities & Activities Report will be invited soon. http://www.haskell.org/communities/ Submissions are welcome between 20th and 31st of October ---------------------------------------------------------------------- The real Call for Contributions will follow closer to the submission window, in about two weeks. This is just an early warning. You will need to plan about 15 minutes for composing your contribution, which would be a brief, ASCII-format, description of your Haskell activities (see earlier editions for plenty of examples). Cheers, Claus (current editor of HC&A Reports)

1 0

Re: ANNOUNCE: HaRe, the Haskell Refactorer, version 0.1
by ketil＠ii.uib.no 02 Oct '03

02 Oct '03

(taken to -cafe) "C.Reinke" <C.Reinke(a)kent.ac.uk> writes: > Dear Haskellers, > > as part of our project on Refactoring Functional Programs > > http://www.cs.kent.ac.uk/projects/refactor-fp/ > > we are pleased to announce the availability of HaRe 0.1 (also > known as HaRe 01/10/2003 ;-), a snapshot of our Haskell Refactorer > prototype. First, let me say that I'm intrigued. This looks like really neat functionality to have available. I'm curious whether you are planning (or have developed) tools to *detect* cases for refactoring? Occasionally, I find myself wishing for a tool to help me clean up a finished program, by e.g. - removing unused items from export/import lists - identifying functions which are only used in one place (and thus candidate for inclusion in 'where' clauses) - identifying functions defined in one module, but only (mainly?) used by/depending on functions in another module, and thus candidate for cross-module migration. - perhaps similarly, identify function transformation that reduces the parameter lists (i.e. exposed function interface) and so on. (Does that make sense at all? :-) -kzm -- If I haven't seen further, it is by standing in the footprints of giants

2 1

AW: Newbie qustion about monads
by Markus.Schnell＠infineon.com 02 Oct '03

02 Oct '03

> I'm not trying to create useful monads (I'm pretty sure they > aren't :), > but understanding the concepts. So, the question remains: > when the monad > laws say that > > (return x) >>= f == f x The Monad class is just called Monad because it is intended to cover a monad. But it doesn't ensure the laws. That is your sole responsibility. > > what is intended in that "=="? Eq."==" for You can define it to your liking as long as it is commutative. HTH, Markus

3 2