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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.
Javier Neira

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

blog.sigfpe.com/...or-people-who-simply-dont.html

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"
J.A. Alonso

A History of Haskell: being lazy with class - 0 views

research.microsoft.com/...index.htm

artículo Haskell historia

shared by J.A. Alonso on 27 Dec 11 - No Cached
  • J.A. Alonso on 27 Dec 11
     
    This long (55-page) paper describes the history of Haskell, including its genesis and principles, technical contributions, implementations and tools, and applications and impact.
Javier Neira

Learning Haskell Notes - 0 views

www.ninebynine.org/...Learning-Haskell-Notes.html

haskell learning notes

shared by Javier Neira on 09 Dec 09 - Cached
  • 8. Functors
  • A "functor" is a structured collection (or container) type with a method (fmap) that accepts a method and applies that method to the members of the collection yielding an isomorphic collection of values of a (possibly) new type. Is this right?
  • Every monad is a functor, but not the other way around; a monad is a functor PLUS functions >>= and return satisfying some laws
  • ...4 more annotations...
  • a functor is a type constructor PLUS a function fmap satisfying some laws.
  • I think it's better to use existentials, as they let you define multiple instances for the same type.
  • People tend to forget that the major difference between ADT's and OO-style classes is really only that with a class you can have many instances in the same program simultaneously, whereas with an ADT you can have only one; but the ADT implementation is still interchangeable.
  • sequence :: Monad m => [m a] -> m [a]
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