Problems with writing a programm in Haskell with signature IO[(Int, Int)] - haskell

I have a real problems with programm, which should be written with signature IO[(Int, Int)].
For me, as really beginner, it's quite hard to understand how it should look like.
So,this is task:
You should write a procedure with takes lines from user until empty line. After that it should return every line length. I try to make it more understandable with example
*Main> Take
kdfdfdf
dfdfeer
ererere
[(7,7),(7,0)]
Reading part it's quite understandable, unless, it's doesn't work actually as I want
Read
= do
putStrLn "User, your turn!"
line <- getLine
if line==""
then return ...
else do
line <-Read
return line {- actually doesn't return a line -}
I will be very glad, if someone make me understand how to write this

What exactly do you want Read to do? As is, it won't compile for a few reasons. You can't name a function or constant starting with a capital letter, those are reserved for Data constructors. Also, I'm sure you know, but the line return ... isn't valid syntax in Haskell.
I don't really understand why you want a list of pairs either if you just want to return the lengths of each line then something like this would work.
lineLengths :: IO [Int]
lineLengths = do
putStrLn "User, your turn!"
line <- getLine
if line == ""
then return []
else do
moreLines <- lineLengths
return $ (length line) : moreLines
This will prompt the user to enter a line, and if it is an empty line then it will return an empty list, assuming you don't care about the length of the empty line required to stop the interaction. Otherwise it will recurse and prepend the length of the line onto the list of lineLengths that was calculated.
EDIT:
If you really want a list of pairs, I think the best way is probably to use another function like this.
toPairs :: [Int] -> [(Int,Int)]
toPairs [] = []
toPairs [x] = [(x,0)]
toPairs (x:y:zs) = (x,y) : toPairs zs
Then to combine this with the reading function we can do this:
lineLengthPairs :: IO [(Int,Int)]
lineLengthPairs = do
ls <- lineLengths
return $ toPairs ls

Related

Reading 2-line input Haskell

I've finished writing a Haskell program but I have troubles with the I/O and Main parts of it. I want to read input of the following format:
10 3
1 4 8
The first and second numbers are Ints, and the numbers of the second line should be made into an integer list, [Int]. The length of the list is equal to the second number on the first line.
I have the following code to read one Int at a time, however, it can only get it to work for the first line of input.
getInt :: IO Int
getInt = do
l <- getLine
return (read l)
What am I doing wrong?
you should be able to get the data shown here by:
readData :: IO (Int, [Int])
readData = do
[nr, _] <- map read . words <$> getLine
nrs <- map read . words <$> getLine
return (nr,nrs)
this will read both lines and return the 10 as the first component and [1,4,8] as the second component of a tuple if you read in your example
As the second number in the first line will just be the length of the returned list I skipped it here.
of course you probably will want to adapt this to your needs.

Reading numbers from input Haskell

I want to have a function that reads arbitrary int's until the number '0' is inserted, and then presents the numbers inserted in an ordered list.
For that i wrote this function:
import Data.List
readIntegers :: IO()
readIntegers = do
putStrLn "insert a number: "
num<-getLine
let list = ordList ((read num :: Int):list)
if (read num == 0)
then print list
else readIntegers
where ordList ::[Int]->[Int]
ordList [] = []
ordList xs = sort xs
This compiles just fine, but when i insert the number '0', it gives me this error:
*** Exception: <<loop>>
What am i doing wrong ?
As #phg points out, you are essentially constructing an infinite list, and actually evaluating it causes the loop error. A simple implementation to resolve this issue is to define a helper function which takes an additional parameter - a list to store all the inputs read in from the screen, like so:
readInteger :: IO ()
readInteger = readInteger' []
where
readInteger' x = do
putStrLn "insert a number: "
num<-getLine
if ((read num :: Int) == 0)
then print $ ordList x
else readInteger' $ (read num :: Int):x
where ordList ::[Int]->[Int]
ordList [] = []
ordList xs = sort xs
Please note that the above is essentially just an implementation of #phg's answer, but with some changes to your original logic. Firstly, since 0 is a sentinel value, we shouldn't be appending that to our list. Second, we do not need to sort the list every single time we are adding a value to it. Sorting once at the time of printing/passing to another function is sufficient.
Demo
If you want to read an unspecified number of integers without prompting for user input and cut it off the moment you encounter 0, you would probably do well to use getContents, which will read everything from the standard input as a single string, lazily.
Then, it is a simple matter of parsing it to a list of numbers and doing what you want with it, like so:
readIntegers :: ()
readIntegers = do
a <- getContents
let b = ordList $ takeWhile (/= 0) $ map (\x -> read x :: Int) $ words a
mapM (putStrLn . show) b
where ordList ::[Int]->[Int]
ordList [] = []
ordList xs = sort xs
let list = ordList ((read num :: Int):list)
This is basically a recursive definition of a list of the form [x, x, ...] (like if you wrote an equation saying x = 1 + x). That is perfectly fine by itself, since Haskell is lazy; however, if you try to print list (aka "solve the equation"), it will fail, since it will try to print infinitely many numbers.
You probably have a misconception about the workings of the (:) operator. Haskell functions will never perform an assignment operation and concatenate num onto list by changing it, like in imperative languages. There are only pure functions.
If you want to accumulate all numbers, you should try to come up with a recursive definition of readIntegers, keeping its state (the list) in an additional parameter (there are also more sophisticated ways, hiding the state passing, but more complicated to use for a beginner).
For a more sophisticated solution, note that this is an unfold and you can use unfoldM from Control.Monad.Loops to implement it:
import Control.Monad.Loops (unfoldM)
readInts :: IO [Int]
readInts = unfoldM $ fmap (check . read) getLine
where check x = if x == 0 then Nothing else Just x
This has the nice property that it returns the list in the order in which it was read.

How to do multiple actions in Haskell

Im trying to write a simple function to learn the IO monad in Haskell. The function is supposed to take the sum of some given integers from the console but when the function has run for example 4 times it says "1*** Exception: Char.digitToInt: not a digit '\n'"
import Data.Char
readInts :: IO ()
readInts = do
putStrLn "Choose nr of ints to sum."
c <- getChar
let i = digitToInt c
-- code.
let sum = getInts i
let str = "\nSum is: " ++ [intToDigit i].
putStrLn str
getInt :: IO Int
getInt = do
c <- getChar
return (digitToInt c)
getInts :: Int -> IO Int
getInts n = if n == 0
then return 0
else
do i <- getInt
j <- getInts
return (i+j)
Can somebody please explain where my recursion is going wrong?
You're simply using the wrong tools to convert between "keyboard data" and numbers. Has little to do with IO.
intTodigit, as the name says, acts on single digits / characters, not on general numbers. What you want is to read / print entire strings, which can handle multi-digit numbers. Replace getChar with getLine, digitToInt with read, and [intToDigit i] with show i. Then it should work fine.
However, it would be better to make some more simplifications.
getInt basically exists already, though in more general form: readLn gets a line from stdin and inteprets it as the required type.
getInts is implemented more complicately than necessary. Explicit recursion over a counting variable (BTW, it has to be getInts (n-1) in the recursion) is ugly; such looping is obviously so common that there exists a standard solution (you need to import Control.Monad) for it which looks alot like loops in imperative languages you might be used to:
getIntsAndSum :: Int -> IO Int
getIntsAndSum n = fmap sum . forM [1 .. n] $ \k -> do
i <- getInt
return i
which can in fact be further simplified to
fmap sum . forM [1 .. n] $ \_ -> getInt
because do blocks a just an optional construct to help you chain actions, but when there's only one action you can simply write that on its own.
Ideally, you would not have to first ask for the number of numbers at all: just gather all numbers you're given, and sum them up. This works, as jamshidh said, very simply with interact.
main = interact processInput
where processInput allInput = show (sum allNumbers)
where allNumbers = map read . lines allInput
and that's it! Nothing else needed. In fact, this can be written yet simpler: you basically have just one simple data-flow pipeline.
main = interact $ show . sum . map read . lines
getChar gets every char, including any "\n" that you type in (which you need to submit the answer). Try filtering these out.
A better way to solve this could be to use "interact" to get the user input, and break apart the data using lines. That way the user could input multi digit numbers, and the "\n"'s would be removed.
The following sums numbers entered at the command line (without any prompting, end inputting by typing ^d).
main = interact (show . sum . fmap read . lines)

Iterating through a String and replacing single chars with substrings in haskell

I am trying to learn some Haskell and I find it difficult. I am having some issues with my
current project. The idea is that I have to go through a String and substitute certain chars
with new substrings. For instance if I have a String "FLXF" and I want to replace every F
with a substring called "FLF" the result should be "FLFLXFLF". Now I have been working on this
specific problem for hours. I have been reading up on types, different functions that might come in handy (map, fold, etc) and yet I have not been able to solve this problem.
The code below is some of the different tries I have had:
apply :: String -> String
apply [] = []
apply (x:xs) = if (x == 'F')
then do show "Hello"
apply xs
else (apply (xs))
This example here I was just trying to show hello every time I encountered a 'F', but all it shows is "", so this clearly does not work. I am really not sure an if else statement is the way to go here. I was also thinking the function map might do the trick. Here the code I was thinking about could look something like this:
map (\x y -> if y == 'F' then "FLD" else y) "FLF"
but that gives me a type error. So as you can see I am lost. Excuse me my poor knowledge to Haskell, but I am still new to it. I really hope some of you can help me out here or give me a push in the right direction. Feel free to ask questions if I have been unclear about something.
Thank you in advance!
John
map (\x y -> if y == 'F' then "FLD" else y) "FLF"
This is nearly right.
First... why does the function take two arguments?
map (\y -> if y == 'F' then "FLD" else y) "FLF"
The remaining type error is because the then branch gives a String, but the else branch gives a Char (the two branches must each give a value of the same type). So we'll make the else branch give a String instead (recall that String is a synonym for [Char]):
map (\y -> if y == 'F' then "FLD" else [y]) "FLF"
Now the problem is that this gives you a [String] value instead of a String. So we'll concatenate all those strings together:
concat (map (\y -> if y == 'F' then "FLD" else [y]) "FLF")
This combination of concat and map is common enough that there's a standard function that combines them.
concatMap (\y -> if y == 'F' then "FLD" else [y]) "FLF"
concatMap is the most intuitive thing here. This kind of combination between mapping over a data structure a function that does itself return the type of the data structure (in this case, a list) and combining the results back into a single "tight" list is indeed very common in Haskell, and indeed not only for lists.
I'd like to explain why your first attempt compiles at all, and what it actually does – because it's completely different from what you probably think!
apply (x:xs) = if (x == 'F')
that line is still perfectly clear: you just take the first char off the string and compare it to 'F'. At bit "pedestrian" to manually take the string apart, but fine. Well, the name you gave the function is not particularly great, but I'll stick with it here.
then do show "Hello"
now this is interesting. You probably think do starts a list of points, "first do this, then do that"... like in simple Hello, World-ish example programs. But always remember: in Haskell, there's normally no such thing as an order in which stuff is calculated. That only happens in the IO context. But there's no IO in your code!?!
Not sure if you've heard about what IO actually is, anyway here you go: it's a Monad. Those "mythical Haskell constructs you've only read about in story books"...
Indeed, though this might lead a bit far here, this question covers all there is to know about Monads! How is that?
Here's another (correct!) way do define your function.
apply' str = do
x <- str
if (x == 'F')
then "FLF"
else return x
So I'm using this weird do syntax, and it's not in IO, and it looks completely different from what you'd write in IO, but it works. How?
x <- str
In do notation, variable <- action always means something like "take one value out of this monadic thingy, and call it x". What you've probably seen is something like
response <- getLine
which means "take a user input out of the real world (out of the IO monad!) and call it response". In x <- str, it's a string that we have, not an IO action. So we take a character out of a string – nice and easy!
Actually, it's not quite right, though. "take a character" is what you do with apply (x:xs) = ..., which simply takes the first one. In contrast, x <- str actually takes all possible characters out of the string, one by one. If you're used to procedural languages, this may seem very inconsistent with response <- getLine, but in fact it's not: getLine also consists of every possible input that the user might give, and the program has to act according to this.
if (x == 'F')
nothing unexpected here, but
then "FLF"
whoah! Just like that? Let's first look at the next line
else return x
ok, this looks familiar, but actually it's not. In other languages, this would mean "we're done with our function, x is the result". But that's obviously not what happens here, because x is Char, and the "return type" of apply' is String. In Haskell, return actually has little to do with returning values from a function, instead it means "put that value into the monadic context that we're working in". If the monad were IO, that would be quite the same: give this value back to the real-world context (this does not mean to print the value or something, just to hand it on). But here, our context is a string, or rather a list (of chars, so it is a String).
Right, so if x is not 'F' we put it back into the string. That sounds reasonable enough, but what about then "FLF"? Note that I can also write it this way:
if (x == 'F')
then do
x' <- "FLF"
return x'
else return x
which means, I take all characters out of "FLW" and return them back into the overall result. But there's no need to only think about the final result, we can as well isolate only this part do { x' <- "FLF"; return x' } – and, quite obviously, its value is nothing but the string "FLF" itself!
So I hope you have now grasped why apply' works. Back to your version, though it actually doesn't make much sense...
then do
show "Hello"
apply xs
here we have a line that's not at the end of a do block, but doesn't have a <- in it. You normally see this in IO in something like
main = do
putStrLn "How ya doin'?"
response <- getLine
...
Remember that "output-only" actions have type IO() in Haskell, which means, they don't directly return any meaningful value, just the trivial value (). So you don't really care about this, but you could still evaluate it:
main = do
trivial <- putStrLn "Hello, let's see what this IO action returns:"
print trivial
compiles and outputs
Hello, let's see what this IO action returns:()
It would be stupid if we had to do this evaluating () all the time, so Haskell allows to just leave the () <- out. It's really just that!
So a line like show "Hello" in the middle of a do block basically means "take one character out of show "Hello" (which is simply a string with the value "\"Hello\""), but don't do anything else with this character / just throw it away".
The rest of your definition is just other recursive calls to apply, but because none of them does anything more interesting than throwing away characters, you eventually end up at apply [] = [], so that's the final result: an empty string.
if-then-else... I know that Haskell supports these, however, I'm very surprised that no one here removed them...
So below are my solutions for different cases of making replacements.
Replacing a character
Replacing words
Replacing through a function on each word
$ cat replace.hs
import Data.List (isPrefixOf)
replaceC :: Char -> Char -> String -> String
replaceC _ _ [] = []
replaceC a b (x:xs)
| x == a = b:replaceC a b xs
| otherwise = x:replaceC a b xs
replaceW :: String -> String -> String -> String
replaceW a b s = unwords . map replaceW' $ words s
where replaceW' x | x == a = b
| otherwise = x
replaceF :: (String -> String) -> String -> String
replaceF f = unwords . map f . words
string = "Hello world ^fg(blue)"
main = do
print string
print $ replaceC 'o' 'z' string
print $ replaceW "world" "kitty" string
print . replaceF f . replaceW "world" "kitty" $ replaceC 'H' 'Y' string
where f s | "^" `isPrefixOf` s = '^':'^':drop 1 s
| otherwise = s
$ runhaskell replace.hs
"Hello world ^fg(blue)"
"Hellz wzrld ^fg(blue)"
"Hello kitty ^fg(blue)"
"Yello kitty ^^fg(blue)"
Your basic error was that you wanted to replace a Char in a String with a String.
This is impossible because String is a list of Char and a Char is a Char and not a short String. Neither is a String ever a Char, even if its length is 1.
Hence, what you really wanted is to replace some Char with some other Chars. Your approach was promising and could have been completed like so:
replace [] = [] -- nothing to replace in an empty string
replace (c:cs) = if c == 'F' then 'F':'L':'F':replace cs
else c:replace cs

haskell front letters

So im educating myself for the future
firstLetter :: IO String
firstLetter = do
x <- getChar
if (x == ' ')
then return (show x)
else firstLetter
So it would get lines until the first line, that starts with empty char
how can I do it, so if empty line comes, it returns all head(x)
for example:
Liquid
Osone
Liquid
(empty line)
returns
"LOL"
Try this. The library function lines will split the input into lines for you, so all that is left is extracting the first character from each string in a list until one string is empty. An empty string is just a null list, so you can check for that to end the recursion over the list of strings.
firstLetters :: [String] -> String
firstLetters (x:xs)
| null x = []
| otherwise = head x : firstLetters xs
main = do
contents <- getContents
putStrLn . firstLetters . lines $ contents
Have you seen interact? This'll help you eliminate the IO and that always seems to make thing simpler for me and hopefully you too.
That reduces it to a problem that reads a string and returns a string.
Here's a rough go at it. getLines takes a string, breaks it into lines and consumes them (takeWhile) until it meets a line containing a single space (I wasn't sure on your ending condition, as the other poster says using null will stop at the first empty list). Then it goes over those lines and gets the first character of each (with map head).
getLines :: String -> String
getLines = map head . takeWhile (/= " ") . lines
main :: IO ()
main = interact getLines

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