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Javier Neira

Monads as containers - HaskellWiki - 0 views

www.haskell.org/...Monads_as_Containers

haskell monads intro as container

shared by Javier Neira on 28 Dec 09 - Cached
  • A monad is a container type together with a few methods defined on it.
  • all the elements which a monadic container holds at any one time must be the same type (it is homogeneous).
  • map (fmap), return and join,
  • ...15 more annotations...
  • map, (but called fmap in Haskell 98) actually comes from the definition of a functor
  • That is, if f is a functor, and we are given a function of type (a -> b), and a container of type (f a), we can get a new container of type (f b). This is expressed in the type of fmap: fmap :: (Functor f) => (a -> b) -> f a -> f b If you will give me a blueberry for each apple I give you (a -> b), and I have a box of apples (f a), then I can get a box of blueberries (f b). Every monad is a functor.
  • The second method, return, is specific to monads. If m is a monad, then return takes an element of type a, and gives a container of type (m a) with that element in it. So, its type in Haskell is return :: (Monad m) => a -> m a If I have an apple (a) then I can put it in a box (m a).
  • takes a container of containers m (m a), and combines them into one m a in some sensible fashion. Its Haskell type is join :: (Monad m) => m (m a) -> m a
  • If I have a box of boxes of apples (m (m a)) then I can take the apples from each, and put them in a new box (m a).
  • bind or extend, which is commonly given the symbol (>>=)
  • liftM :: (Monad m) => (a -> b) -> m a -> m b liftM f xs = xs >>= (return . f) -- take a container full of a's, to each, apply f, -- put the resulting value of type b in a new container, -- and then join all the containers together.
  • The function that does this for any monad in Haskell is called liftM -- it can be written in terms of return and bind as follows:
  • Well, in Haskell, IO is a monad.
  • Lists are most likely the simplest, most illustrative example
  • The reason that bind is so important is that it serves to chain computations on monadic containers together.
  • You might notice a similarity here between bind and function application or composition, and this is no coincidence.
  • What bind does is to take a container of type (m a) and a function of type (a -> m b). It first maps the function over the container, (which would give an m (m b)) and then applies join to the result to get a container of type (m b). Its type and definition in Haskell is the
  • xs >>= f = join (fmap f xs)
  • bind (>>=)
Javier Neira

What is (functional) reactive programming? - Stack Overflow - 0 views

stackoverflow.com/...1028250

reactive programming functional definition stackoverflow

shared by Javier Neira on 07 Jan 10 - Cached
  • Semantically, FRP's concurrency is fine-grained, determinate, and continuous.
  • Dynamic/evolving values (i.e., values "over time") are first class values in themselves. You can define them and combine them, pass them into & out of functions. I called these things "behaviors".
  • Each occurrence has an associated time and value.
  • ...3 more annotations...
  • In software design, I always ask the same question: "what does it mean?".
  • It's been quite a challenge to implement this model correctly and efficiently, but that's another story.
  • The basic idea behind reactive programming is that there are certain datatypes that represent a value "over time". Computations that involve these changing-over-time values will themselves have values that change over time.
Javier Neira

Haskell: The Confusing Parts - 0 views

echo.rsmw.net/n00bfaq.html

haskell intro good syntax

shared by Javier Neira on 15 Nov 09 - Cached
  • f $ x = f x(f . g) x = f (g x)-- Some quick examples of using these. The following line...putStrLn (take 12 (map foo (bar ++ "ack")))-- ...can be rewritten as...putStrLn $ take 12 $ map foo $ bar ++ "ack"(putStrLn . take 12 . map foo) (bar ++ "ack")putStrLn . take 12 . map foo $ bar ++ "ack" In terms of good Haskell style, the last example above is preferable to the others. By the way, note that ($) has the lowest precedence (zero) of any operator, so you can almost always use arbitrary syntax on either side of it.
  • -- Actual definitions:f $ x = f x(f . g) x = f (g x)-- Some quick examples of using these. The following line...putStrLn (take 12 (map foo (bar ++ "ack")))-- ...can be rewritten as...putStrLn $ take 12 $ map foo $ bar ++ "ack"(putStrLn . take 12 . map foo) (bar ++ "ack")putStrLn . take 12 . map foo $ bar ++ "ack"
  • putStrLn (take 12 (map foo (bar ++ "ack")))-- ...can be rewritten as...putStrLn $ take 12 $ map foo $ bar ++ "ack"(putStrLn . take 12 . map foo) (bar ++ "ack")putStrLn . take 12 . map foo $ bar ++ "ack"
  • ...7 more annotations...
  • putStrLn (take 12 (map foo (bar ++ "ack")))
  • anywhere you might write \x -> foo x [1..10], you should instead write flip foo [1..10], where flip is a standard Prelude function that flips the first two arguments of whatever function you give it
  • two arguments of whatever
  • anywhere you might write \x -> foo x [1..10], you should instead write flip foo [1..10], where flip is a standard Prelude function that flips the first
  • The curry and uncurry functions stand in for \f x y -> f (x, y) and \f (x, y) -> f x y, respectively.
  • The curry and uncurry functions stand in for \f x y -> f (x, y)
  • I will point out that return is, in fact, not a return statement. It’s a function, and an inappropriately named function, at that. Writing return () in your do block will not cause the function to return.
  • Javier Neira on 15 Nov 09
     
    If you're used to the C family of languages, or the closely related family of "scripting languages," Haskell's syntax (mainly) is a bit baffling at first. For some people, it can even seem like it's sneaking out from under you every time you think you understand it. This is sort of a FAQ for people who are new to Haskell, or scared away by its syntax.
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