Re: [Haskell-cafe] Monad transformer, liftIO
Creighton Hogg schrieb: > Okay, so I think what you want is > > [...] Yes. Your solution works. Thank you. But: > a <- msum . map return $ [1,2,3] Why Do I need this "msum . map return" thing? The "map return" part is somewhat clear. But not entirely. Which type of monad is created here? The msum-part ist totally confusing me: First we create a list with some monads (?) and then msum them? What is going on there? > first, that you tried to use literal list syntax in > do notation which I believe only works in the actual [] monad. Is the "x <- xs" in the list monad a form of syntactic sugar? > Second, you didn't have the runListT acting on > the foobar, which is how you go from a ListT IO Int to a IO [Int]. Ah, yes. This point I understood now. Thank you again. Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Monad transformer, liftIO
Hello list, maybe I'm just stupid, I'm trying to do something like this: import Control.Monad import Control.Monad.Trans import Control.Monad.List foobar = do a <- [1,2,3] b <- [4,5,6] liftIO $ putStrLn $ (show a) ++ " " ++ (show b) return (a+b) main = do sums <- foobar print sums But this apparently doesn't work... I'm total clueless how to achieve the correct solution. Maybe my mental image on the monad transformer thing is totally wrong? Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Field updates in a state monad
Hello list,
still playing with monads and states, I have the following question:
Given:
import Control.Monad.State.Lazy
data MyData = MyData { content :: String }
foobar :: State MyData String
foobar = do
gets content
Ok, that looks nice and tidy. But:
foobar2 :: State MyData ()
foobar2 = do
modify $ \x -> x { content = "hello haskell"}
...looks not so nice.
Exists there a way to write this cleaner without writing countless
"set_xyz" helper functions?
Michael
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Re: [Haskell-cafe] Problem with own written monad
Jules Bean schrieb:
> data Stack a b = Stack { run :: [a] -> (b, [a]) }
Thank you, that does the trick.
> The correct types for the other functions are:
>
> push :: a -> Stack a ()
> pop :: Stack a a
> top :: Stack a a
>
> With those clues I think you will be able to write >>= and return more
> successfully!
Yes, this was the missing link. Because I thought "Stack a a" could be
abbreviated using "Stack a" I run into these problems. This was also the
cause that "push" echoed back the pushed value.
> There are some other interesting combinators to consider, like:
>
> isolate :: Stack b x -> Stack a x
> -- runs the computation with an empty stack, therefore
> -- guaranteeing it does more pushes than pops
Did you mean:
isolate :: Stack s1 a -> Stack s2 a
isolate stack = Stack f where f xs = ( fst $ run stack [], xs)
> and so on.
Yes, I have done: push, pop, top, nop, count, clear, isolate and binop.
All pretty easy, once I understand that "Stack a b" thing.
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Re: [Haskell-cafe] Problem with own written monad
Miguel Mitrofanov schrieb: > May be you can explain what do you want to do with this "monad"? Pure educational purpose, just "learning by doing". > What kind of code would you write if it would be such monad? Useless stuff like: s2 = do push 11 push 17 count >>= push binop (+) binop (*) pop ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Problem with own written monad
Hello list,
while trying to learn the secrets of monads, I decided to write a simply
monand for pure educational purpose. But it turned out that it isn't as
easy as I thought... I circumnavigate quite a number of hurdles but now
I reached a point where I'm at a loss. :-(
The source:
#! /usr/bin/env ghc
data Stack a = Stack { run :: [a] -> (a, [a]) }
push :: a -> Stack a
push x = Stack f where f xs = ( x, x:xs )
pop :: Stack a
pop = Stack f where f (x:xs) = ( x, xs )
top :: Stack a
top = Stack f where f (x:xs) = ( x, x:xs )
instance Monad Stack where
return x = Stack f where f xs = ( x, xs )
(>>=) stack g = Stack f where
f s0 = (x2, s2) where
(x1, s1) = run stack s0
(x2, s2) = run (g x1) s1
The errors:
./mymonad.hs:16:24:
Couldn't match expected type `b' (a rigid variable)
against inferred type `a' (a rigid variable)
`b' is bound by the type signature for `>>=' at
`a' is bound by the type signature for `>>=' at
Expected type: [b] -> (b, [b])
Inferred type: [a] -> (b, [b])
In the first argument of `Stack', namely `f'
In the expression: Stack f
./mymonad.hs:19:28:
Couldn't match expected type `b' (a rigid variable)
against inferred type `a' (a rigid variable)
`b' is bound by the type signature for `>>=' at
`a' is bound by the type signature for `>>=' at
Expected type: [b]
Inferred type: [a]
In the second argument of `run', namely `s1'
In the expression: run (g x1) s1
I think the problem is that my operator (>>=) is of type:
Stack a -> (a -> Stack a) -> Stack a
but should be:
Stack a -> (a -> Stack b) -> Stack b
But, I have simply no clue how to fix that. :-(
Can anybody give my a hint?
Thank you in advance.
Michael Roth
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[Haskell-cafe] How to write elegant Haskell programms? (long posting)
Hello list,
I'm new to Haskell and I'm trying to learn how to write elegant code
using Haskell.
I decided to convert the following small tool, written in ruby:
===
#! /usr/bin/env ruby
require 'pathname'
BASENAMES = %w{ mail.log thttpd.log }
ARCHIVEDIR = Pathname.new '/var/log/archive'
LOGDIR = Pathname.new '/var/log'
class Pathname
def glob glob_pattern
Pathname.glob self.join(glob_pattern)
end
def timestamp
stat.mtime.strftime '%Y%m%d'
end
end
for basename in BASENAMES
for oldname in LOGDIR.glob "#{basename}.*.gz"
newname = ARCHIVEDIR.join "#{basename}.#{oldname.timestamp}.gz"
puts "mv #{oldname} #{newname}"
File.rename oldname, newname
end
end
===
My solution in Haskell is:
===
import System.Directory (getDirectoryContents, getModificationTime,
renameFile)
import System.Locale (defaultTimeLocale)
import System.Time(ClockTime, toUTCTime, formatCalendarTime)
import Text.Regex (mkRegex, matchRegex)
import Maybe
import Control.Monad
logdir, archivedir :: String
logfiles :: [String]
logfiles= [ "mail.log", "thttpd.log" ]
logdir = "/var/log"
archivedir = "/var/log/archive"
basename :: String -> String
basename filename = head . fromMaybe [""] $ matchRegex rx filename where
rx = mkRegex "^(.+)(\\.[0-9]+\\.gz)$"
isLogfile :: String -> Bool
isLogfile filename = basename filename `elem` logfiles
timestamp :: ClockTime -> String
timestamp time =
formatCalendarTime defaultTimeLocale "%Y%m%d" (toUTCTime time)
makeOldname :: String -> String
makeOldname fn = logdir ++ '/' : fn
makeNewname :: String -> String -> String
makeNewname bn ts = archivedir ++ '/' : bn ++ '.' : ts ++ ".gz"
move :: String -> String -> IO ()
move oldname newname = do
putStrLn $ "mv " ++ oldname ++ ' ' : newname
renameFile oldname newname
main :: IO ()
main = do
files <- liftM (filter isLogfile) (getDirectoryContents logdir)
let oldnames = map makeOldname files
times <- mapM getModificationTime oldnames
let newnames = zipWith makeNewname (map basename files) (map timestamp
times)
zipWithM_ move oldnames newnames
===
Ok, the tool written in Haskell works. But, to me, the source doesn't
look very nice and even it is larger than the ruby solution, and more
imporant, the programm flow feels (at least to me) not very clear.
Are there any libraries available to make writing such tools easier?
How can I made the haskell source looking more beautiful?
Michael Roth
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