haskell evaluation $ sign - haskell

I am going through 'learn you some haskell' and I have written following application:
import System.IO
main = do
filename <- getLine
handle <- openFile filename ReadMode
content <- hGetContents handle
putStr . unlines . (map isLong) . lines $ content
hClose handle
isLong :: String -> String
isLong x = if length x > 10 then x ++ " long enough" else x ++ " could be better!"
And it works but when I remove "$" between lines and content a compilation fails.
Could you help me understand why this is wrong?
I was thinking that I compose statements with dots and I get a function (String -> IO ()) and I apply it to the "content" but why is "$" needed here?.
Thanks!

The operator (.) has type (b -> c) -> (a -> b) -> a -> c.... Its first two inputs must be functions.
lines content, however, is of type [String], not a function, hence f . lines content will fail. The compiler treats it as
f . (lines content)
By adding the ($), you change the precedence, and it becomes
f . lines $ content = (f . lines) $ content
which works, because f and lines are both functions.

The dollar sign in haskell is used for application of a function on a value.
Its backround is, that you do not need complicated parentheses in terms.

Related

How to get this function to be evaluated lazily

I have the following function:
main = do xs <- getContents
edLines <- ed $ lines xs
putStr $ unlines edLines
Firstly I used the working version main = interact (unlines . ed . lines) but changed the signature of ed since. Now it returns IO [String] instead of just [String] so I can't use this convenient definition any more.
The problem is that now my function ed is still getting evaluated partly but nothing is displayed till I close the stdin via CTRL + D.
Definition of ed:
ed :: Bool -> [EdCmdLine] -> IO EdLines
ed xs = concatM $ map toLinesExt $ scanl (flip $ edLine defHs) (return [Leaf ""]) xs where
toLinesExt :: IO [EdState] -> IO EdLines
toLinesExt rsIO = do
rs#(r:_) <- rsIO -- todo add fallback pattern with (error)
return $ fromEd r ++ [" "]
The scanl is definitely evaluated lazy because edLine is getting evaluated for sure (observable by the side effects).
I think it could have to do with concatM:
concatM :: (Foldable t, Monad m) => t (m [a]) -> m [a]
concatM xsIO = foldr (\accIO xIO -> do {x <- xIO; acc <- accIO; return $ acc ++ x}) (return []) xsIO
All I/O in Haskell is explicitly ordered. The last two lines of your main function desugar into something like
ed (lines xs) >>= (\edLines -> putStr $ unlines edLines)
>>= sequences all of the I/O effects on the left before all of those on the right. You're constructing an I/O action of the form generate line 1 >> ... >> generate line n >> output line 1 >> ... >> output line n.
This isn't really an evaluation order issue, it's a correctness issue. An implementation is free to evaluate in any order it wants, but it can't change the ordering of I/O actions that you specified, any more than it can reorder the elements of a list.
Here's a toy example showing what you need to do:
lineProducingActions :: [IO String]
lineProducingActions = replicate 10 getLine
wrongOrder, correctOrder :: IO ()
wrongOrder = do
xs <- sequence lineProducingActions
mapM_ putStrLn xs
correctOrder = do
let xs = [x >>= putStrLn | x <- lineProducingActions]
sequence_ xs
Note that you can decouple the producer and consumer while getting the ordering you want. You just need to avoid combining the I/O actions in the producer. I/O actions are pure values that can be manipulated just like any other values. They aren't side-effectful expressions that happen immediately as they're written. They happen, rather, in whatever order you glue them together in.
You would need to use unsafeInterleaveIO to schedule some of your IO actions for later. Beware that the IO actions may then be executed in a different order than you might first expect!
However, I strongly recommend not doing that. Change your IO [String] action to print each line as it's produced instead.
Alternately, if you really want to maintain the computation-as-pipeline view, check out one of the many streaming libraries available on Hackage (streamly, pipes, iteratees, conduit, machines, and probably half a dozen others).
Thanks to #benrg answer I was able to solve the issue with the following code:
ed :: [EdCmdLine] -> [IO EdLines]
ed cmds = map (>>= return . toLines . head) $ edHistIO where
toLines :: EdState -> EdLines
toLines r = fromEd r ++ [" "]
edHistIO = edRec defHs cmds (return [initState])
edRec :: [HandleHandler] -> [EdCmdLine] -> IO EdHistory -> [IO EdHistory]
edRec _ [] hist = [hist] -- if CTRL + D
edRec defHs (cmd:cmds) hist = let next = edLine defHs cmd hist in next : edRec defHs cmds next
main = getContents >>= mapM_ (>>= (putStr . unlines)) . ed . lines

Reading numbers inline

Imagine I read an input block via stdin that looks like this:
3
12
16
19
The first number is the number of following rows. I have to process these numbers via a function and report the results separated by a space.
So I wrote this main function:
main = do
num <- readLn
putStrLn $ intercalate " " [ show $ myFunc $ read getLine | c <- [1..num]]
Of course that function doesn't compile because of the read getLine.
But what is the correct (read: the Haskell way) way to do this properly? Is it even possible to write this function as a one-liner?
Is it even possible to write this function as a one-liner?
Well, it is, and it's kind of concise, but see for yourself:
main = interact $ unwords . map (show . myFunc . read) . drop 1 . lines
So, how does this work?
interact :: (String -> String) -> IO () takes all contents from STDIN, passes it through the given function, and prints the output.
We use unwords . map (show . myFunc . read) . drop 1 . lines :: String -> String:
lines :: String -> [String] breaks a string at line ends.
drop 1 removes the first line, as we don't actually need the number of lines.
map (show . myFunc . read) converts each String to the correct type, uses myFunc, and then converts it back to a `String.
unwords is basically the same as intercalate " ".
However, keep in mind that interact isn't very GHCi friendly.
You can build a list of monadic actions with <$> (or fmap) and execute them all with sequence.
λ intercalate " " <$> sequence [show . (2*) . read <$> getLine | _ <- [1..4]]
1
2
3
4
"2 4 6 8"
Is it even possible to write this function as a one-liner?
Sure, but there is a problem with the last line of your main function. Because you're trying to apply intercalate " " to
[ show $ myFunc $ read getLine | c <- [1..num]]
I'm guessing you expect the latter to have type [String], but it is in fact not a well-typed expression. How can that be fixed? Let's first define
getOneInt :: IO Int
getOneInt = read <$> getLine
for convenience (we'll be using it multiple times in our code). Now, what you meant is probably something like
[ show . myFunc <$> getOneInt | c <- [1..num]]
which, if the type of myFunc aligns with the rest, has type [IO String]. You can then pass that to sequence in order to get a value of type IO [String] instead. Finally, you can "pass" that (using =<<) to
putStrLn . intercalate " "
in order to get the desired one-liner:
import Control.Monad ( replicateM )
import Data.List ( intercalate )
main :: IO ()
main = do
num <- getOneInt
putStrLn . intercalate " " =<< sequence [ show . myFunc <$> getOneInt | c <- [1..num]]
where
myFunc = (* 3) -- for example
getOneInt :: IO Int
getOneInt = read <$> getLine
In GHCi:
λ> main
3
45
23
1
135 69 3
Is the code idiomatic and readable, though? Not so much, in my opinion...
[...] what is the correct (read: the Haskell way) way to do this properly?
There is no "correct" way of doing it, but the following just feels more natural and readable to me:
import Control.Monad ( replicateM )
import Data.List ( intercalate )
main :: IO ()
main = do
n <- getOneInt
ns <- replicateM n getOneInt
putStrLn $ intercalate " " $ map (show . myFunc) ns
where
myFunc = (* 3) -- replace by your own function
getOneInt :: IO Int
getOneInt = read <$> getLine
Alternatively, if you want to eschew the do notation:
main =
getOneInt >>=
flip replicateM getOneInt >>=
putStrLn . intercalate " " . map (show . myFunc)
where
myFunc = (* 3) -- replace by your own function

Reading multiline user's input

I want to lazily read user input and do something with it line by line. But if user ends a line with , (comma) followed by any number of spaces (including zero), I want give him opportunity to finish his input on the next line.
And here is what I've got:
import System.IO
import Data.Char
chop :: String -> [String]
chop = f . map (++ "\n") . lines
where f [] = []
f [x] = [x]
f (x : y : xs) = if (p . tr) x
then f ((x ++ y) : xs)
else x : f (y : xs)
p x = (not . null) x && ((== ',') . last) x
tr xs | all isSpace xs = ""
tr (x : xs) = x :tr xs
main :: IO ()
main =
do putStrLn "Welcome to hell, version 0.1.3!"
putPrompt
mapM_ process . takeWhile (/= "quit\n") . chop =<< getContents
where process str = putStr str >> putPrompt
putPrompt = putStr ">>> " >> hFlush stdout
Sorry, it doesn't work at all. Bloody mess.
P.S. I want to preserve \n characters on end of every chunk. Currently I add them manually with map (++ "\n") after lines.
How about changing the type of chop a little:
readMultiLine :: IO [String]
readMultiLine = do
ln <- getLine
if (endswith (rstrip ln) ",") then
liftM (ln:) readMultiLine
else
return [ln]
Now you know that if the last list is not empty, then the user didn't finish typing (the last input ended with ',').
Of course, either import Data.String.Utils, or write your own. Could be as simple as:
endswith xs ys = (length xs >= length ys)
&& (and $ zipWith (==) (reverse xs) (reverse ys))
rstrip = reverse . dropWhile isSpace . reverse
But I missed the point at first. Here's the actual thing.
unfoldM :: (Monad m) => (a -> Maybe (m b, m a)) -> a -> m [b]
unfoldM f z = case f z of
Nothing -> return []
Just (x, y) -> liftM2 (:) x $ y >>= unfoldM f
main = unfoldM (\x -> if (x == ["quit"]) then Nothing
else Just (print x, readMultiLine)) =<< readMultiLine
The reason is, you need to be able to insert the "action" to be done on input between reading one multi-line input and the next. Here print x is the action inserted between two readMultiLine
Since you have questions about getContents, let me add. Even though getContents provides a lazy String, its effectful changes to the world are ordered with the subsequent effects of processing the list. But the processing of the list attempts to insert effects between effects of reading particular list items. To do that, you need a function that exposes the chain of effects, so you can insert your own effects between them.
You can do this using pipes, preserving the laziness of the user's input
import Data.Char (isSpace)
import Pipes
import qualified Pipes.Prelude as Pipes
endsWithComma :: String -> Bool
endsWithComma str =
case (dropWhile isSpace $ reverse str) of
',':_ -> True
_ -> False
finish :: Monad m => Pipe String String m ()
finish = do
str <- await
yield str
if endsWithComma str
then do
str' <- await
yield str'
else finish
user :: Producer String IO ()
user = Pipes.stdinLn >-> finish
You can then hook up the user Producer to any downstream Consumer. For example, to echo the stream back out you can write:
main = runEffect (user >-> Pipes.stdoutLn)
To learn more about pipes you can read the tutorial.
Sorry, I wrote something wrong in a comment and I thought that now that I understood what you were trying to do, I'd give an answer with a little more substance. The core idea is that you're going to need a state buffer while you loop through the string, as far as I can tell. You have f :: [String] -> [String] but you'll need an extra string of buffer before you can solve this puzzle.
So let me assume an answer which looks like:
chop = joinCommas "" . map (++ "\n") . lines
Then the structure of joinCommas is going to look like:
import Data.List (isSuffixOf)
-- override with however you want to handle the ",\n" between lines.
joinLines = (++)
incomplete = isSuffixOf ",\n"
joinCommas :: String -> [String] -> [String]
joinCommas prefix (line : rest)
| incomplete prefix = joinCommas (joinLines prefix line) rest
| otherwise = prefix : joinCommas line rest
joinCommas prefix []
| incomplete prefix = error "Incomplete input"
| otherwise = [prefix]
The prefix stores up lines until it doesn't end with ",\n" at which point it emits the prefix and continues with the rest of the lines. On EOF we process the last line unless that line is incomplete.

Non-exhaustive patterns in lambda

I am getting Non-exhaustive patterns in lambda. I am not sure of the cause yet. Please anyone how to fix it. The code is below:
import Control.Monad
import Data.List
time_spent h1 h2 = max (abs (fst h1 - fst h2)) (abs (snd h1 - snd h2))
meeting_point xs = foldl' (find_min_time) maxBound xs
where
time_to_point p = foldl' (\tacc p' -> tacc + (time_spent p p')) 0 xs
find_min_time min_time p = let x = time_to_point p in if x < min_time then x else min_time
main = do
n <- readLn :: IO Int
points <- fmap (map (\[x,y] -> (x,y)) . map (map (read :: String->Int)) . map words . lines) getContents
putStrLn $ show $ meeting_point points
This is the lambda with the non-exhaustive patterns: \[x,y] -> (x,y).
The non-exhaustive pattern is because the argument you've specified, [x,y] doesn't match any possible list - it only matches lists with precisely two elements.
I would suggest replacing it with a separate function with an error case to print out the unexpected data in an error message so you can debug further, e.g.:
f [x,y] = (x, y)
f l = error $ "Unexpected list: " ++ show l
...
points <- fmap (map f . map ...)
As an addition to #GaneshSittampalam's answer, you could also do this with more graceful error handling using the Maybe monad, the mapM function from Control.Monad, and readMaybe from Text.Read. I would also recommend refactoring your code so that the parsing is its own function, it makes your main function much cleaner and easier to debug.
import Control.Monad (mapM)
import Text.Read (readMaybe)
toPoint :: [a] -> Maybe (a, a)
toPoint [x, y] = Just (x, y)
toPoint _ = Nothing
This is just a simple pattern matching function that returns Nothing if it gets a list with length not 2. Otherwise it turns it into a 2-tuple and wraps it in Just.
parseData :: String -> Maybe [(Int, Int)]
parseData text = do
-- returns Nothing if a non-Int is encountered
values <- mapM (mapM readMaybe . words) . lines $ text
-- returns Nothing if a line doesn't have exactly 2 values
mapM toPoint values
Your parsing can actually be simplified significantly by using mapM and readMaybe. The type of readMaybe is Read a => String -> Maybe a, and in this case since we've specified the type of parseData to return Maybe [(Int, Int)], the compiler can infer that readMaybe should have the local type of String -> Maybe Int. We still use lines and words in the same way, but now since we use mapM the type of the right hand side of the <- is Maybe [[Int]], so the type of values is [[Int]]. What mapM also does for us is if any of those actions fails, the overall computation exits early with Nothing. Then we simply use mapM toPoint to convert values into a list of points, but also with the failure mechanism built in. We actually could use the more general signature of parseData :: Read a => String -> Maybe [(a, a)], but it isn't necessary.
main = do
n <- readLn :: IO Int
points <- fmap parseData getContents
case points of
Just ps -> print $ meeting_point ps
Nothing -> putStrLn "Invalid data!"
Now we just use fmap parseData on getContents, making points have the type Maybe [(Int, Int)]. Finally, we pattern match on points to print out the result of the meeting_point computation or print a helpful message if something went wrong.
If you wanted even better error handling, you could leverage the Either monad in a similar fashion:
toPoint :: [a] -> Either String (a, a)
toPoint [x, y] = Right (x, y)
toPoint _ = Left "Invalid number of points"
readEither :: Read a => String -> Either String a
readEither text = maybe (Left $ "Invalid parse: " ++ text) Right $ readMaybe text
-- default value ^ Wraps output on success ^
-- Same definition with different type signature and `readEither`
parseData :: String -> Either String [(Int, Int)]
parseData text = do
values <- mapM (mapM readEither . words) . lines $ text
mapM toPoint values
main = do
points <- fmap parseData getContents
case points of
Right ps -> print $ meeting_point ps
Left err -> putStrLn $ "Error: " ++ err

Haskell, can i call function without IO output working with monads?

Why i can't do this?
Its forbidden the use of 'do' in this question :/
How i can call words in my list and at same time result an IO?
Thanks.. this is my actual code :/
main :: IO()
main =
putStr "Name of File: " >>
getLine >>=
\st ->
openFile st ReadMode >>=
\handle ->
hGetContents handle >>=
\y ->
words y >>=
\strings ->
strings !! 1 >>=
\string->
putStr string
[Edit] Solution :
main :: IO()
main =
putStr "Name of File: " >>
getLine >>=
\st ->
openFile st ReadMode >>=
\handle ->
hGetContents handle >>=
\y ->
return (words y) >>=
\strings ->
return (strings !! 1) >>=
\string->
putStr string
Use return (words y) instead of just words y. return wraps a pure value (such as the [String] that words returns) into a monad.
From your wording, it sounds like this question is homework. If so, it should be tagged as such.
(This doesn't directly answer the question, but it will make your code more idiomatic and thus easier to read.)
You are using the pattern \x -> f x >>= ... a lot, this can (and should) be eliminated: it is (mostly) unnecessary noise which obscures the meaning of the code. I won't use your code, since it is homework but consider this example (note that I'm using return as suggested by the other answer):
main = getLine >>=
\fname -> openFile fname ReadMode >>=
\handle -> hGetContents handle >>=
\str -> return (lines str) >>=
\lns -> return (length lns) >>=
\num -> print num
(It reads a file name from the user, and then prints the number of lines in that file.)
The easiest optimisation is the section where we count the number of lines (this corresponds to the part where you are separating the words and getting the second one): the number of lines in a string str is just length (lines str) (which is the same as length . lines $ str), so there is no reason for us to have the call to length and the call to lines separate. Our code is now:
main = getLine >>=
\fname -> openFile fname ReadMode >>=
\handle -> hGetContents handle >>=
\str -> return (length . lines $ str) >>=
\num -> print num
Now, the next optimisation is on \num -> print num. This can be written as just print. (This is called eta conversion). (You can think about this as "a function that takes an argument and calls print on it, is the same as print itself"). Now we have:
main = getLine >>=
\fname -> openFile fname ReadMode >>=
\handle -> hGetContents handle >>=
\str -> return (length . lines $ str) >>= print
The next optimisation we can do is based on the monad laws. Using the first one, we can turn return (length . lines $ str) >>= print into print (length . lines $ str) (i.e. "creating a container that contains a value (this is done by return) and then passing that value to print is the same as just passing the value to print"). Again, we can remove the parenthesis, so we have:
main = getLine >>=
\fname -> openFile fname ReadMode >>=
\handle -> hGetContents handle >>=
\str -> print . length . lines $ str
And look! We have an eta-conversion we can do: \str -> print . length . lines $ str becomes just print . length . lines. This leaves:
main = getLine >>=
\fname -> openFile fname ReadMode >>=
\handle -> hGetContents handle >>= print . length . lines
At this point, we can probably stop, since that expression is much simpler than our original one (we could keep going, by using >=> if we wanted to). Since it is so much simpler, it is also easier to debug (imagine if we had forgotten to use lines: in the original main it wouldn't be very clear, in the last one it's obvious.)
In your code, you can and should do the same: you can use things like sections (which mean \x -> x !! 1 is the same as (!! 1)), and the eta conversion and monad laws I used above.

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