When I use this code it's print newline after result. How I can don't write newline?
import System.IO
main :: IO ()
main = do
a <- getLine
b <- getLine
let aa = read a :: Int
let bb = read b :: Int
let cc = aa + bb
print cc
print is defined as putStrLn . show, the fix is to use putStr . show $ cc.
You can also putStr instead of putStrLn. Not sure how you would do something similar with print though.
Related
I need help in coding this because I keep getting this error:
practice.hs:44:40: error: parse error on input `='
|
44 | sortedList :: [Integer] sortedList = sort numList
|
I hope someone can help. Here is my source code
import Data.List
listBuilder :: IO()
listBuilder = do
numList []
main = do
listBuilder
putStrLn "Enter 5 Numbers"
num1 <- getLine
num2 <- getLine
num3 <- getLine
num4 <- getLine
num5 <- getLine
num1 :: numList
num2 :: numList
num3 :: numList
num4 :: numList
num5 :: numList
sortedList :: [Integer] sortedList = sort numList
print ("Old List: " ++ numList)
print ("Sorted List: " ++ sortedList)```
First off, there are no "variables" in Haskell - values are immutable!
Secondly, :: Something is a type annotation, so num1 :: numList would mean "num1 is of type numList" - but what is "numList" supposed to be?
Here is a very simple solution:
import Data.List
main = do
putStrLn "Enter 5 Numbers"
line1 <- getLine
let num1 = read line1 :: Int
line2 <- getLine
let num2 = read line2 :: Int
line3 <- getLine
let num3 = read line3 :: Int
line4 <- getLine
let num4 = read line4 :: Int
line5 <- getLine
let num5 = read line5 :: Int
let list = [num1, num2, num3, num4, num5]
let sortedList = sort list
putStr "Old List: "
print list
putStr "Sorted List: "
print sortedList
OK, so I read a line from standard in, using getLine. This has type IO String, so it is an IO action that yields a String.
Using read, which has type Read a => String -> a, we turn the String into an Int. The type means "some 'a' which has an instance of typeclass Read, take a string and spit out an 'a', if possible".
I'm using a type annotation :: Int to signal, that I want that a to be an Int.
Repeating that 5 times, I end up with 5 Ints.
I create a list using the [] list constructor (note: you can use : [read "cons"] to put an element to the front of the linked list).
sort from Data.List will sort the list in ascending order, using the compare operation. (There is also sortOn, so you can provide a function to sort the elements by).
Then I use putStr to print my string (without a newline).
The print function takes an a and calls show on it, to turn it into a string, before printing it. So you don't need to use print with strings (you can use putStrLn (newline at end) or putStr (no newline)).
This is getting into advanced territory, but the solution is not elegant and we can do better!
Basically, we want to repeat the same IO action 5 times and convert a list of strings to a list of Ints:
import Data.List
import Control.Monad
main = do
putStrLn "Enter 5 Numbers"
inputs <- replicateM 5 getLine
let list = map read inputs :: [Int]
let sortedList = sort list
putStr "Old List: "
print list
putStr "Sorted List: "
print sortedList
repeatM repeats a monadic action n times.
The <- means something like "take this value out of the monadic context, so IO [String] and we want the [String] part.
map just applies the provided function (here read) to every element of the list argument.
I'm using a type annotation again, to tell Haskell, that the result of this expression should be a list of Ints (otherwise read would not know what a is supposed to be).
I know what an indentation error is, but I have no idea why I'm getting this error here, while every is aligned, trying to solve it for 2 hours.
Account.hs:40:25: error:
parse error (possibly incorrect indentation or mismatched brackets)
|
40 | let amount = readLn :: IO Int
| ^
Failed, 0 modules loaded.
main = do
putStrLn $ "Press one to create a new account"
let g = getLine
enteredValue = read g :: Int
if g == 1
then do putStrLn $ "Enter your name "
let name = getLine
putStrLn $ "Enter the initial amount"
let amount = readLn :: IO Int
value = Account (name,1,amount) Saving
show value
else do putStrLn $ "Nothing"
I also tried this version but this also gives incorrect indentation or mismatched brackets:
main = do
putStrLn $ "Press one to create a new account"
let g = getLine
enteredValue = read g :: Int
if g == 1
then do putStrLn $ "Enter your name "
let name = getLine
putStrLn $ "Enter the initial amount"
amount = readLn :: IO Int
value = Account (name,1,amount) Saving
show value
else do putStrLn $ "Nothing"
The problem is here:
-- |<---- "column 0" of this 'do' block
then do putStrLn $ "Enter your name "
-- | still good; a 'let' statement:
let name = getLine
-- |<---- "column 0" of this 'let' block
putStrLn $ "Enter the initial amount"
-- | Huh, there's no '=' in ^this^ declaration?
let amount = readLn :: IO Int
-- ^^^ Why is there a 'let' within another let binding?
-- I still haven't seen a '='. Better throw a parse error.
Basically, putStrLn $ "Enter the initial amount" is aligned with name = ... in the preceding line, so the compiler reads it as a declaration (part of the same let block).
To fix your indentation errors, it should be:
main = do
putStrLn $ "Press one to create a new account"
let g = getLine
enteredValue = read g :: Int
if g == 1
then do putStrLn $ "Enter your name "
let name = getLine
putStrLn $ "Enter the initial amount"
let amount = readLn :: IO Int
value = Account (name,1,amount) Saving
show value
else do putStrLn $ "Nothing"
But then you'll run into type errors:
read g is wrong: read takes a String, but g :: IO String
g == 1 is wrong: 1 is an Int, but g :: IO String
show value is wrong: show returns a String, but you're using it as an IO action
You haven't shown the declaration of Account, but you're likely going to have issues with name and amount, too
You probably want something like:
main = do
putStrLn $ "Press one to create a new account"
g <- getLine
let enteredValue = read g :: Int
if enteredValue == 1
then do putStrLn $ "Enter your name "
name <- getLine
putStrLn $ "Enter the initial amount"
amount <- readLn :: IO Int
let value = Account (name,1,amount) Saving
putStrLn (show value)
else do putStrLn $ "Nothing"
Basically, use v <- expr to go from expr :: IO Something to v :: Something.
Other notes:
g <- getLine; let enteredValue = read g :: Int better written as enteredValue <- readLn :: IO Int
putStrLn (show value) can be shortened to print value
you don't need do for a single expression (nor $ for a single operand): ... else putStrLn "Nothing"
There is more wrong to your code than just the Indentation Errors - so my first suggestion would be reading a bit of learn you a haskell for great good.
Next there are two assignment operators in haskell - one binds the result of an action … <- … and the other one is a local definition/declaration of a pure computation let … = ….
Moreover you can improve your reading a value by taking account of the possible false input, that someone could give you (intentionally and unintentionally) by replacing read with readMaybe, where the latter returns a Maybe something, for example readMaybe "1" = Just 1 :: Maybe Int or readMaybe "foo" = Nothing :: Maybe Int.
Regarding your indentation it is best that you compare one solution to your program with yours own:
import Text.Read (readMaybe)
data Type = Saving | Checking
deriving (Show)
data Account = Account (String,Int,Int) Type
deriving (Show)
main :: IO ()
main = do
putStrLn "Press one to create a new account"
g <- getLine
let enteredValue = readMaybe g :: Maybe Int
here the result of getLine and entered value have the same scope so they have the same indentation - we only change the scope after the next if where the then-block - and the else-block do not share the 'declarations' of each branch, so you couldn't use name in the else-block, but enteredValue can be used in both.
if enteredValue == Just 1
then do putStrLn "Enter your name "
name <- getLine
putStrLn "Enter the initial amount"
amount' <- fmap readMaybe getLine
here again name and amount' share the same scope, and pattern matching on amount' creates a new scope where amount is visible and the match on Nothing where you cannot use this variable.
case amount' of
Just amount -> print $ Account (name,1,amount) Saving
Nothing -> putStrLn "Nothing"
else putStrLn "Nothing"
let is for binding values, which is done in the form let x = y+z, where x is the name (aka "identifier") being bound, and y+z is the expression to which it is being bound.
In your example, I see three bindings: name, amount, and value. The rest are not value bindings, but actions.
In the do notation, actions do not need a let. You just write them one after another. So:
let name = getLine
putStrLn $ "Enter the initial amount"
let amount = readLn :: IO Int
let value = Account (name,1,amount) Saving
show value
But wait! This is not all!
getLine is not actually an expression of type String, as you seem to be hoping here. Rather, getLine is an action. In order to get it to "run" and "produce" a String value, you need to use the <- construct instead of let:
name <- getLine
Similarly with readLn:
amount <- readLn :: IO Int
Finally, show value is not actually an action that would print the value to the screen. show is a function that takes a value and return a String. It doesn't "do" anything (i.e. doesn't produce any outside effects), so you can't use it in place of an action in the do notation. If you wanted an action that would print a value to the screen, that would be print:
print value
Gathering everything together:
name <- getLine
putStrLn $ "Enter the initial amount"
amount <- readLn :: IO Int
let value = Account (name,1,amount) Saving
print value
And after fixing all of that, you'll have similar difficulties with the first part of your program, where you have let g = getLine instead of g <- getLine.
I have a file in the form:
3
1 2
3 4
5 7
Where the first line is the number of lines
I know that:
getInt :: IO Int
getInt = readLn
main = do num <- getInt
print (num)
Reads the first line.
Next, I tried:
readInts :: IO [Int]
readInts = fmap (map read.words) getLine
For read a line and get a list: [a,b].
And I tried to use the above in a recursive loop
loop :: Int -> IO()
loop n = if 1 == n then do num <- readInts
print(num)
else loop (n-1)
I'm getting the first line only:
[5,3]
But I need to read the rest of lines, given N
The format of the input file looks a lot like the ones used in programming contests. Here is my standard setup for programming contests like that:
import Control.Monad
import Text.Printf
main :: IO ()
main = do
n <- readLn
forM_ [1 .. n] $ \i -> do
printf "Case %d: " (i :: Int)
solve
An example of solve might be:
solve :: IO ()
solve = do
nums <- map read . words <$> getLine
print (sum nums)
Adding to this helpful answer and comment, for some challenges you'll need to collect n lines and then emit a single result at the end based on an aggregation of the data. An approach for creating a list might use replicateM as follows:
import Control.Monad
toInt :: String -> Int
toInt x = read x :: Int
lineToInts :: String -> [Int]
lineToInts x = map toInt $ words x
main :: IO ()
main = do
n <- readLn
remainingLines <- replicateM n getLine
let list = map lineToInts remainingLines
print list
Sample run:
3
0 1
3 4
6 8
[[0,1],[3,4],[6,8]]
See also Read n lines input with Haskell
Have you looked into the function lines? It takes a string and returns the same string as a list separated by \n. Using this function you don't even have to have the number of lines.
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
This code reads the number of lines to process from the first line of stdin, then it loops number_of_lines_to_process times doing some calculations and prints the result.
I want it to print the line number in "Line #" after "#" but I don't know how to obtain it
import IO
import Control.Monad (replicateM)
main :: IO ()
main = do
hSetBuffering stdin LineBuffering
s <- getLine
let number_of_lines_to_process = read s :: Integer
lines <- replicateM (fromIntegral(number_of_lines_to_process)) $ do
line <- getLine
let number = read line :: Integer
result = number*2 --example
putStrLn ("Line #"++": "++(show result)) --I want to print the number of the iteration and the result
return ()
I guess that the solution to this problem is really easy, but I'm not familiar with Haskell (coding in it for the first time) and I didn't find any way of doing this. Can anyone help?
You could use forM_ instead of replicateM:
import IO
import Control.Monad
main :: IO ()
main = do
hSetBuffering stdin LineBuffering
s <- getLine
let number_of_lines_to_process = read s :: Integer
forM_ [1..number_of_lines_to_process] (\i -> do
line <- getLine
let number = read line :: Integer
result = number * 2
putStrLn $ "Line #" ++ show i ++ ": " ++ show result)
Note that because you use forM_ (which discards the results of each iteration) you don't need the additional return () at the end - the do block returns the value of the last statement, which in this case is the () which is returned by forM_.
The trick is to first create a list of all the line numbers you want to print, and to then loop through that list, printing each number in turn. So, like this:
import Control.Monad
import System.IO
main :: IO ()
main = do
hSetBuffering stdin LineBuffering
s <- getLine
let lineCount = read s :: Int
-- Create a list of the line numbers
lineNumbers = [1..lineCount]
-- `forM_` is like a "for-loop"; it takes each element in a list and performs
-- an action function that takes the element as a parameter
forM_ lineNumbers $ \ lineNumber -> do
line <- getLine
let number = read line :: Integer
result = number*2 --example
putStrLn $ "Line #" ++ show lineNumber ++ ": " ++ show result
return ()
Read the definition of forM_.
By the way, I wouldn't recommend using the old Haskell98 IO library. Use System.IO instead.
You could calculate the results, enumerate them, and then print them:
import IO
import Control.Monad (replicateM)
-- I'm assuming you start counting from zero
enumerate xs = zip [0..] xs
main :: IO ()
main = do
hSetBuffering stdin LineBuffering
s <- getLine
let number_of_lines_to_process = read s :: Integer
lines <- replicateM (fromIntegral(number_of_lines_to_process)) $ do
line <- getLine
let number = read line :: Integer
result = number*2 --example
return result
mapM_ putStrLn [ "Line "++show i++": "++show l | (i,l) <- enumerate lines ]
I'm still new at Haskell, so there could be problems with the program below (it does work). This program is a tail recursive implementation. The doLine helper function carries around the line number. The processing step is factored into process, which you can change according to the problem you are presented.
import System.IO
import Text.Printf
main = do
hSetBuffering stdin LineBuffering
s <- getLine
let number_of_lines_to_process = read s :: Integer
processLines number_of_lines_to_process
return ()
-- This reads "max" lines from stdin, processing each line and
-- printing the result.
processLines :: Integer -> IO ()
processLines max = doLine 0
where doLine i
| i == max = return ()
| otherwise =
do
line <- getLine
let result = process line
Text.Printf.printf "Line #%d: %d\n" (i + 1) result
doLine (i + 1)
-- Just an example. (This doubles the input.)
process :: [Char] -> Integer
process line = let number = read line :: Integer
in
number * 2
I'm a haskell rookie, so any critiques of the above are welcome.
Just as an alternative, I thought that you might enjoy an answer with minimal monad mucking and no do notation. We zip a lazy list of the user's data with an infinite list of the line number using the enumerate function to give us our desired output.
import System.IO
import Control.Monad (liftM)
--Here's the function that does what you really want with the data
example = (* 2)
--Enumerate takes a function, a line number, and a line of input and returns
--an ennumerated line number of the function performed on the data
enumerate :: (Show a, Show b, Read a) => (a->b) -> Integer -> String -> String
enumerate f i x = "Line #" ++
show i ++
": " ++
(show . f . read $ x) -- show . f . read handles our string conversion
-- Runover takes a list of lines and runs
-- an enumerated version of the sample over those lines.
-- The first line is the number of lines to process.
runOver :: [String] -> [String]
runOver (line:lines) = take (read line) $ --We only want to process the number of lines given in the first line
zipWith (enumerate example) [1..] lines -- run the enumerated example
-- over the list of numbers and the list of lines
-- In our main, we'll use liftM to lift our functions into the IO Monad
main = liftM (runOver . lines) getContents