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mesbah095

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haskell del.icio.us imported Tutorial functional programming reference intro monads howto

shared by mesbah095 on 03 Apr 14 - No Cached
  • mesbah095 on 03 Apr 14
     
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Javier Neira

Applicative-what? Functor-who? « wxfz :: Blog - 0 views

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haskell control structures functors arrows monads intro

shared by Javier Neira on 09 Sep 10 - No Cached
  • Monads Are  a kind of abstract data type used to represent computations (instead of data in the domain model).
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

Learn You a Haskell for Great Good! - Types and Typeclasses - 0 views

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type classes haskell tutorial intro

shared by Javier Neira on 22 Dec 09 - Cached
  • That's why we can use explicit type annotations. Type annotations are a way of explicitly saying what the type of an expression should be. We do that by adding :: at the end of the expression and then specifying a type.
Javier Neira

A Neighborhood of Infinity: The IO Monad for People who Simply Don't Care - 0 views

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haskell io monad basics intro

shared by Javier Neira on 22 Dec 09 - No Cached
  • Many programming languages make a distinction between expressions and commands.
  • Like other languages it makes the distinction, and like other languages it has its own slightly idiosyncratic notion of what the difference is. The IO monad is just a device to help make this distinction.
  • There is no room for anything like a print command here because a print command doesn't return a value, it produces output as a side-effect
  • ...18 more annotations...
  • It's easy to use: you just write do and then follow it by an indented list of commands. Here's a complete Haskell program:
  • Note also that commands take arguments that can be expressions. So print (2*x) is a command, but 2*x is an expression. Again, little different from a language like Python.
  • So here's an interesting feature of Haskell: commands can return values. But a command that returns a value is different from an expression with a value.
  • We have to use <- instead of let ... = ... to get a returned value out of a command. It's a pretty simple rule, the only hassle is you have to remember what's a command and what's a function.
  • get2Lines = do line1 <- getLine line2 <- getLine return (line1,line2)
  • To introduce a list of commands, use do.To return a value from a command, use return.To assign a name to an expression inside a do block use let.To assign a name to the result of a command, use <-.
  • what's a command and what's an expression? If it has any chance of doing I/O it must be a command, otherwise it's probably an expression.
  • Everything in Haskell has a type, even commands. In general, if a command returns a value of type a then the command itself is of type IO a.
  • eturn is simply a function of type a -> IO a that converts a value of type a to a command of type IO a.
  • 5. The type of a sequence of commands is the type of the last line.
  • The type of an if statement is the type of its branches. So if you want an if statement inside a do block, it needs to be a command, and so its branches need to be commands also. So it's
  • If something is of type IO a then it's a command returning an value of type a. Otherwise it's an expression. That's the rule.
  • following the rules above it's completely impossible to put an executed command inside an expression.
  • As the only way to do I/O is with commands, that means you have no way to find out what Haskell is doing inside expressions.
  • If the type isn't IO a, then you can sleep at night in the confidence that there are no side effects.
  • One last thing. Much of what I've said above is false. But what I hope I have done is describe a subset of Haskell in which you can start some I/O programming without having a clue what a monad is.
  • The idea of capturing imperative computations in a type of (immutable) values is lovely. And so is the general pattern we call "monad".
  • main = do return 1 print "Hello"
Javier Neira

Haskell: The Confusing Parts - 0 views

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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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