June 2016 - Beginners

How to hunt this space-leak
by martin 13 Jun '16

13 Jun '16

Hello all, I hope I am not asking for too much, as to explain my problem, you need to read my code, which may not be a pleasure. I am trying to write a "Logger", which formats and accumulates log-entries in a monoid. From there I went to writing an "accumulating" logger, i.e. one which can accumulate (sum/avg) over entries made with a certain timespan. My problen is memory consumption. In the test below I stress the logger with 1 million values, but in the end result, there are only 10 entries left, because it accumulates over 100,000. Memory goes up to apx 100MB. When I comment out the line, which logs the accumulated value (see -- > below), memory stays below 10MB. I dont understand why this is so. After all, the difference is only whether or not those 10 entries are logged or not. Can someone explain this? {-# LANGUAGE BangPatterns#-} import Data.Monoid import Control.Monad.State.Strict import System.TimeIt import Text.Show.Pretty import Debug.Trace ------------------------------------------------------------ -- Simple Time Stuff ------------------------------------------------------------ type Instant = Double type Interval = Double type Timed a = (Instant, a) instant = fst ------------------------------------------------------------ -- Logger based on State monad ------------------------------------------------------------ data SLogger a l = SLgr {runSLogger :: a ->State l (SLogger a l)} type SLogFormatter a l = a -> l accLogger :: (Monoid c, Show a) => (Instant, Interval, [Timed b]) -> SLogFormatter (Timed a) [Timed b] -> SLogFormatter [Timed b] [Timed c] -> SLogger (Timed a) [Timed c] accLogger (tx, dt, tas) fmt1 fmt2 = SLgr $ \(!ta) -> let x = fmt1 ta !tas' = x <> tas in if instant ta < tx then do -- keep accumulating return $ accLogger (tx, dt, tas') fmt1 fmt2 else do -- compute new log and reset accumulator !l0 <- get -- > put $ fmt2 tas' <> l0 return $ accLogger ((tx+dt), dt, []) fmt1 fmt2 accFmt1 ta = [ta] accFmt2 tas = [(fst $ head tas, "hello from accFormatter")] -- apply logger to a list of as stest lgr [] = return lgr stest lgr (a:as) = do lgr' <- (runSLogger lgr) a stest lgr' as main2 = do let as = zip [1.0 .. 1000000.0] [1..1000000] :: [(Instant, Int)] log = execState (stest (accLogger (100000.0,100000.0,[]) accFmt1 accFmt2 ) as) [(0,"init")] timeIt $ putStrLn $ ppShow log putStrLn "done" main = main2

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Haskell and design patterns
by Dennis Raddle 09 Jun '16

09 Jun '16

I have been using Haskell on and off, as a hobbyist, for five years. I haven't been diligent in training myself on it, so I am learning slowly. It was confusing at first, coming from an imperative background. It occurred to me the other day that Haskell typeclasses are a kind of "design pattern." They identify a common pattern and provide a solution. I've been reading about design patterns in C++ for years. Haskell patterns are a twist on that. The imperative view looks at a task in a certain way, often conceiving of it as steps that mimic how human thoughts would go, and finding big general patterns that occur in different big applications. I don't want to speak for all of Haskell, but at least some Haskell patterns are smaller-scale and more attuned to the language of programming rather than the language of human thought. For instance, the Monad. The common use case is this---you have some "things," and you have a "container" that can accept more than one type of "thing." The common patterns are - you need to wrap things in the container - you need to unwrap things - you need to modify things while in their container - you have functions that operate on containers - you have functions that operate on the things, but maybe you don't want to unwrap them first. or maybe you do. To an imperative programmer, this is bizarre. What things are you talking about? What containers? You have not defined either one, so why can you say anything specific? Why all this emphasis on wrapping and unwrapping? Yet this is a design pattern all the same. It gets weird when you realize that the Maybe monad, conceived in human-thought terms, has totally different behavior than the list monad. Yet they are the same design pattern and have a relationship that is more concisely expressed mathematically than in human thought. The abundance of operators for wrapping/unwrapping/operating seems, if I understand right, closely related to the conciseness of Haskell. If you can use your containers without unwrapping them first, then by all means you don't want extra syntax just to do so. Same if you have functions that only operate on unwrapped things. Haskell patterns are "lightweight" also helping with the conciseness. The closest thing to a "container" in C++ is a class, which is a dozen times heavier than a Functor, Applicative, or Monad. Actually you really need a "template class" which is like a tank compared to Functor which is a skinny guy fighting with a short knife (who can usually get the job done just fine, thank you). Any comments welcome. I'm just trying to get a good perspective on Haskell. Mike

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Re: [Haskell-beginners] Functions as containers; implications on being a Monad ((->) r)
by Alex Belanger 04 Jun '16

04 Jun '16

Does it help if I bring your: (\x -> x+7) <$> (+3) => (+10) It doesn't magically become `(+10)`. The previous step unseen is `(+7) . (+3)`. fmap for ((->) r) is defined as function composition `.`. fmap f g = f . g You can think of functors as containers or contexts; both analogies work well up to some degree (with the latter maybe more abstract and thus flexible). Another small terminology tip that could help you with your thinkering might be to call ((->) r) a partially applied function. Suddently, it narrows how you can potentially use/combine them. Last tip, Monad ((->) r) is commonly refered to as the Reader monad. Example of how it'd be used: doThings = do a <- (*2) b <- (+10) return (a + b) Notice how we can work with partially applied functions to which something have yet to be applied and create a new partial function whose argument is going to fill all the open holes. This isn't easy to explain without looking at the types of everything, so I encourage you to write it down and work out the types (: Alex On Jun 4, 2016 12:41 AM, "Raja" <rajasharan(a)gmail.com> wrote: Everyone agrees ((->) r) is a Functor, an Applicative and a Monad but I've never seen any good writeup going into details of explaining this. So I was trying to brainstorm with my brother and went pretty far into the concept for quite a few hours, but still got stuck when it came to Monads. Before I showcase the question/problem I wanted to share our thinking process. Lets stick with common types like Maybe a, [a], simple function (a -> b) **Everything is a Container** Just 4 => this is a container wrapping some value [1,2,3] => this is a container wrapping bunch of values (+3) => this is a container wrapping domains & ranges (infinite dictionary) **When is a Container a Functor** If we can peek inside a container and apply a simple function (a->b) to each of its values and wrap the result back inside the container, then it becomes a Functor. Let's use (\x -> x+7) as simple function along with above three Containers (\x -> x+7) <$> Just 4 => Just 11 (\x -> x+7) <$> [1,2,3] => [8,9,10] (\x -> x+7) <$> (+3) => (+10) -- well there is no Show defined but you get the idea **When is a Container an Applicative** The simple function from above is also now wrapped inside a container and we should be able to peek to use it just like functor. Also lets simplify (\x -> x+7) to (+7). Just (+7) <*> Just 4 => Just 11 [(+7)] <*> [1,2,3] => [8,9,10] (\x -> (+7)) <*> (+3) => (+10) -- again no Show defined but works when pass a number -- but (+7) still needs to be wrapped in a Container though **When is a Container a Monad** This time we don't have a simple function (a->b) instead we have a non-simple function (a -> Container). But rest stays almost the same. We have to first peek inside a container and apply non-simple function to each of its values. Since its a non-simple function we need to collect all the returned Containers, unwrap them and wrap them all back in a Container again. (it almost feels like unwrap and wrapping them back is going to complicate things) Also Non-simple function cannot be reused as is for all three Containers like in Functors & Applicatives. Just 4 >>= (\x -> Just (x+7)) => Just 11 [1,2,3] >>= (\x -> [x+7]) => [8,9,10] (+3) >>= (\x -> (+7)) => (+7) Wait a minute ... the last line doesn't look right. Or should I say it doesn't feel right to discard the `x' altogether. OK let's jump to do this: (+3) >>= (\x -> (+x)) => ??? -- apparently this solves for the equation: f(x) = 2x + 3 (is 2x + 3 obvious for anyone??? it took us way longer to derive it) (Does it have anything to do with Monad laws by any chance?) This is where it feels like "Functions as containers" concept starts to breakdown; its not the right analogy for Monads. What does it even mean to unwrap a bunch of functions and wrap them back again? Hope this intrigues some of you as it did to us. Any thoughts and comments greatly appreciated. Thanks, Raja _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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help on writing a typeclass
by Silent Leaf 04 Jun '16

04 Jun '16

I got a function, named mdk (don't ask). It happens it's a method of a class, as follows: class C a where mdk :: C b => a -> b -> (a, b) yet i got another function, called so far mdk' mdk' :: C b => b -> a -> (a, b) Say there are the following instances: instance C X where mdk x b = ... :: C b => (X, b) mdk' b x = ... :: C b => (X, b) instance C Y where mdk y b = ... :: C b => (Y, b) mdk' b y = ... :: C b => (Y, b) if you get what's happening, mdk and mdk' are basically the same method, save for the final type-result, which is reversed, without data loss mind you (aka (A, B) and (B, A) could be the same type, if only i knew of a way to make a type in haskell without forcing the order (a kind of set or something). so, there are four cases: input have types, in this order, X and Y, and expected output is (X, Y): then mdk from X's instance of C is called/must be used; input have types Y and X, output (X, Y): X's mdk' input X and Y, output (Y, X): Y's mdk input Y and X, output (Y, X): Y's mdk' it might seem confusing, but all amounts to the issue that even though (Y, X) is strictly equivalent, in my program, to (X, Y), in principle haskell differentiate the types. thus i have to write two versions of mdk for each type, even though the methods "X's mdk" and "Y's mdk'" are identical, and same for the two other methods. indeed: mdk :: C b => X -> b -> (X, b) mdk' :: C b => b -> Y -> (Y, b) are identical signatures, if in the first one you replace `b` with Y, and in the latter you replace `b` with X, and ofc if you consider (X,Y) == (Y,X). only one of those two methods should be enough to handle the job, and be chosen by the compiler on the sole value of the type of the first variable; but the fact both methods return tuples of "mirrored" types, crushes that. if i only write `mdk` instances, as soon as the compiler will meet this following signature: mdk :: a -> b -> (b, a) it will crash an exception, because mdk's original signature has for output value a tuple whose first type should here be `a`, because it's meant to be the type of the first argument, not the second. thus, at last my question: can i tell the compiler to consider (b, a) and (a, b) as identical types (don't worry it's not really a tuple in my program, but it's equivalent)? if not, can i make an overloading of mdk so it accepts both a->b->(a,b) and a->b->(b,a)? hope i didn't lose anyone. if so, do tell me, i'll try to clarify. thanks in advance of the time spent trying to understand my problem!

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"type" declarations: utterly transparent synonyms?
by Silent Leaf 03 Jun '16

03 Jun '16

I'd like to be sure when one defines a type with "type", it truly merely creates a synonym, and no matter what one could replace the left side by the right side in the whole program, and vice versa, and it wouldn't change a thing in the actual meaning, value of the program. Eg if i define: type In2 t = t -> t -> t I can thereafter replace for any type t, any signature like f :: t -> t -> t with f :: In2 t then strictly no consequence other than visual in the source code, by me, will occur. Thanks!

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why is there no typeclass common to all containers?
by Silent Leaf 03 Jun '16

03 Jun '16

All in the title. I haven't used them much, but I saw Map or Vector types were forcing the user to use qualified functions unless you want nameclash with the more basic, typically list-oriented functions. So, why not have a massive, general purpose interface so the type only can separate between containers --which would allow for cross-container polymorphism, i suppose, more easily, even though it's not necessarily the most widespread need. So, do i miss something? Is there in fact a class of that kind? If so why not? Thanks in advance! :)

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