I have the following function:
get :: Chars -> IO Chars
get cs = do
char <- getChar
let (dats, idx) = (curData cs, curIndex cs)
let (x,y:xs) = splitAt idx dats
let replacement = x ++ (ord char) : xs
return $ Chars replacement idx
and I'd like to get a Chars value out of it, not an IO action. I have no idea how to do this, or if it is even possible.
The Chars type is basically just a container:
data Chars = Chars {
curData :: [Int],
curIndex :: Int
-- etc.
}
The specifics aren't that important, I just want to know if there's a way for this function to return a Chars instead of an IO Chars.
If not, how do I pass this as an argument to a function that takes a Chars? I'm kind of new to Haskell I/O, but I don't think I want all of my functions that take Chars as arguments to instead have to take IO Chars as arguments, and then extract and repackage them. It seems unnecessary.
Thanks!
You can't, because that would violate referential transparency.
IO in Haskell is made this way exactly to distinguish between actions whose result and effects may vary depending on the interaction with the environment/user and pure functions whose results are not going to change when you call them with the same input parameters.
In order to pass the result to a pure function taking a Chars in input you have to call your IO action into another IO action, bind the result with the <- operator to a variable and pass it to your pure function. Pseudocode example:
myPureFunction :: Chars -> ...
otherAction :: Chars -> IO ()
otherAction cs = do
myChars <- get cs
let pureResult = myPureFunction myChars
...
If you're new to IO in haskell, you may wish to have a look at the Input and Output chapters in Learn You a Haskell for a Great Good! and Real World Haskell.
There is actually a way to simply get a pure value out of an IO action, but in your case you shouldn't do it, as you're interacting with the environment: the unsafe way is ok only when you can guarantee you're not violating referential transparency.
It's impossible (I lie, there is an extremely unsafe way to cheat your way out of it).
The point is that if any I/O is performed, the behaviour and result of your programme may not depend only on explicit arguments to the used functions, thus that must be declared in the type by having it IO something.
You use the result of an IO a action in a pure function by binding the result in main (or something called from main) and then applying the pure function, binding the result in a let,
cs ::Chars
cs = undefined
main = do
chars <- get cs
let result = pureFunction chars
print result
or, if the function you want to apply to chars has type Chars -> IO b
main = do
chars <- get cs
doSomething chars
Related
This question already has answers here:
How to get normal value from IO action in Haskell
(2 answers)
Closed 1 year ago.
In many imperative programming languages like Java, C or Python we can easily add a print function which can give us information about the intermediate state of the program.
My goal is to find a way to do something like that in Haskell. I want the function which not only computes value but also prints something. The function below is a simplified version of what I want to do. My actual function is too complicated and incomprehensive without context to show it here.
My idea is to have a "pure" Haskell function that has an auxiliary function inside which has [Int] -> IO () -> Int type signature. An IO parameter is initialized in the where clause as a do block. But unfortunately, the do block is not executed, when I run the function in GHCI. The function is compiled successfuly though
module Tests where
-- Function returns the sum of the list and tries to print some info
-- but no IO actually happens
pureFuncWithIO :: [Int] -> Int
pureFuncWithIO [] = 0
pureFuncWithIO nums = auxIOfunc nums (return ())
where
auxIOfunc [] _ = 0
auxIOfunc (n : ns) _ = n + auxIOfunc ns (sneakyIOaction n)
sneakyIOaction n
= do -- Not executed
putStrLn $ "adding " ++ (show n);
return ()
Output in GHCI test:
*Tests> pureFuncWithIO [1,2,3,4,5]
15
Meanwhile, I expected something like this:
*Tests> pureFuncWithIO [1,2,3,4,5]
adding 1
adding 2
adding 3
adding 4
adding 5
15
Is it possible to come up with a way to have IO inside, keeping the return type of the outer-most function, not an IO a flavor? Thanks!
This type signature
pureFuncWithIO :: [Int] -> Int
is promising to the caller that no side effect (like prints) will be observed. The compiler will reject any attempt to perform IO. Some exceptions exist for debugging (Debug.Trace), but they are not meant to be left in production code. There also are some "forbidden", unsafe low-level functions which should never be used in regular code -- you should pretend these do not exist at all.
If you want to do IO, you need an IO return type.
pureFuncWithIO :: [Int] -> IO Int
Doing so allows to weave side effects with the rest of the code.
pureFuncWithIO [] = return 0
pureFuncWithIO (n : ns) = do
putStrLn $ "adding " ++ show n
res <- pureFuncWithIO ns
return (n + res)
A major point in the design of Haskell is to have a strict separation of functions which can not do IO and those who can. Doing IO in a non-IO context is what the Haskell type system was designed to prevent.
Your sneakyIOaction is not executed because you pass its result as a parameter to auxIOfunc, but never use that parameter, and haskell being lazy bastard it is never execute it.
If you try to use said parameter you find out that you can't. It's type not allow you to do anithing with it except combine with other IO things.
There is a way to do what you want, but it is on dark side. You need unsafePerformIO
unsafePerformIO :: IO a -> a
That stuff basically allow you to execute any IO. Tricky thing you have to consume result, otherwise you may end up with haskell skip it due to its laziness. You may want to look into seq if you really want to use it, but don't actually need result.
Take the function getLine - it has the type:
getLine :: IO String
How do I extract the String from this IO action?
More generally, how do I convert this:
IO a
to this:
a
If this is not possible, then why can't I do it?
In Haskell, when you want to work with a value that is "trapped" in IO, you don't take the value out of IO. Instead, you put the operation you want to perform into IO, as well!
For example, suppose you want to check how many characters the getLine :: IO String will produce, using the length function from Prelude.
There exists a helper function called fmap which, when specialized to IO, has the type:
fmap :: (a -> b) -> IO a -> IO b
It takes a function that works on "pure" values not trapped in IO, and gives you a function that works with values that are trapped in IO. This means that the code
fmap length getLine :: IO Int
represents an IO action that reads a line from console and then gives you its length.
<$> is an infix synonym for fmap that can make things simpler. This is equivalent to the above code:
length <$> getLine
Now, sometimes the operation you want to perform with the IO-trapped value itself returns an IO-trapped value. Simple example: you wan to write back the string you have just read using putStrLn :: String -> IO ().
In that case, fmap is not enough. You need to use the (>>=) operator, which, when specialiced to IO, has the type IO a -> (a -> IO b) -> IO b. In out case:
getLine >>= putStrLn :: IO ()
Using (>>=) to chain IO actions has an imperative, sequential flavor. There is a kind of syntactic sugar called "do-notation" which helps to write sequential operation like these in a more natural way:
do line <- getLine
putStrLn line
Notice that the <- here is not an operator, but part of the syntactic sugar provided by the do notation.
Not going into any details, if you're in a do block, you can (informally/inaccurately) consider <- as getting the value out of the IO.
For example, the following function takes a line from getLine, and passes it to a pure function that just takes a String
main = do
line <- getLine
putStrLn (wrap line)
wrap :: String -> String
wrap line = "'" ++ line ++ "'"
If you compile this as wrap, and on the command line run
echo "Hello" | wrap
you should see
'Hello'
If you know C then consider the question "How can I get the string from gets?" An IO String is not some string that's made hard to get to, it's a procedure that can return a string - like reading from a network or stdin. You want to run the procedure to obtain a string.
A common way to run IO actions in a sequence is do notation:
main = do
someString <- getLine
-- someString :: String
print someString
In the above you run the getLine operation to obtain a String value then use the value however you wish.
So "generally", it's unclear why you think you need a function of this type and in this case it makes all the difference.
It should be noted for completeness that it is possible. There indeed exists a function of type IO a -> a in the base library called unsafePerformIO.
But the unsafe part is there for a reason. There are few situations where its usage would be considered justified. It's an escape hatch to be used with great caution - most of the time you will let monsters in instead of letting yourself out.
Why can't you normally go from IO a to a? Well at the very least it allows you to break the rules by having a seemingly pure function that is not pure at all - ouch! If it were a common practice to do this the type signatures and all the work done by the compiler to verify them would make no sense at all. All the correctness guarantees would go out of the window.
Haskell is, partly, interesting precisely because this is (normally) impossible.
For how to approach your getLine problem in particular see the other answers.
Consider the code below taken from a working example I've built to help me learn Haskell. This code parses a CSV file containing stock quotes downloaded from Yahoo into a nice simple list of bars with which I can then work.
My question: how can I write a function that will take a file name as its parameter and return an OHLCBarList so that the first four lines inside main can be properly encapsulated?
In other words, how can I implement (without getting all sorts of errors about IO stuff) the function whose type would be
getBarsFromFile :: Filename -> OHLCBarList
so that the grunt work that was being done in the first four lines of main can be properly encapsulated?
I've tried to do this myself but with my limited Haskell knowledge, I'm failing miserably.
import qualified Data.ByteString as BS
type Filename = String
getContentsOfFile :: Filename -> IO BS.ByteString
barParser :: Parser Bar
barParser = do
time <- timeParser
char ','
open <- double
char ','
high <- double
char ','
low <- double
char ','
close <- double
char ','
volume <- decimal
char ','
return $ Bar Bar1Day time open high low close volume
type OHLCBar = (UTCTime, Double, Double, Double, Double)
type OHLCBarList = [OHLCBar]
barsToBarList :: [Either String Bar] -> OHLCBarList
main :: IO ()
main = do
contents :: C.ByteString <- getContentsOfFile "PriceData/Daily/yhoo1.csv" --PriceData/Daily/Yhoo.csv"
let lineList :: [C.ByteString] = C.lines contents -- Break the contents into a list of lines
let bars :: [Either String Bar] = map (parseOnly barParser) lineList -- Using the attoparsec
let ohlcBarList :: OHLCBarList = barsToBarList bars -- Now I have a nice simple list of tuples with which to work
--- Now I can do simple operations like
print $ ohlcBarList !! 0
If you really want your function to have type Filename -> OHLCBarList, it can't be done.* Reading the contents of a file is an IO operation, and Haskell's IO monad is specifically designed so that values in the IO monad can never leave. If this restriction were broken, it would (in general) mess with a lot of things. Instead of doing this, you have two options: make the type of getBarsFromFile be Filename -> IO OHLCBarList — thus essentially copying the first four lines of main — or write a function with type C.ByteString -> OHLCBarList that the output of getContentsOfFile can be piped through to encapsulate lines 2 through 4 of main.
* Technically, it can be done, but you really, really, really shouldn't even try, especially if you're new to Haskell.
Others have explained that the correct type of your function has to be Filename -> IO OHLCBarList, I'd like to try and give you some insight as to why the compiler imposes this draconian measure on you.
Imperative programming is all about managing state: "do certain operations to certain bits of memory in sequence". When they grow large, procedural programs become brittle; we need a way of limiting the scope of state changes. OO programs encapsulate state in classes but the paradigm is not fundamentally different: you can call the same method twice and get different results. The output of the method depends on the (hidden) state of the object.
Functional programming goes all the way and bans mutable state entirely. A Haskell function, when called with certain inputs, will always produce the same output. Simple examples of
pure functions are mathematical operators like + and *, or most of the list-processing functions like map. Pure functions are all about the inputs and outputs, not managing internal state.
This allows the compiler to be very smart in optimising your program (for example, it can safely collapse duplicated code for you), and helps the programmer not to make mistakes: you can't put the system in an invalid state if there is none! We like pure functions.
The exception to the rule is IO. Code that performs IO is impure by definition: you could call getLine a hundred times and never get the same result, because it depends on what the user typed. Haskell handles this using the type system: all impure functions are marred with the IO type. IO can be thought of as a dependency on the state of the real world, sort of like World -> (NewWorld, a)
To summarise: pure functions are good because they are easy to reason about; this is why Haskell makes functions pure by default. Any impure code has to be labelled as such with an IO type signature; this tells the compiler and the reader to be careful with this function. So your function which reads from a file (a fundamentally impure action) but returns a pure value can't exist.
Addendum in response to your comment
You can still write pure functions to operate on data that was obtained impurely. Consider the following straw-man:
main :: IO ()
main = do
putStrLn "Enter the numbers you want me to process, separated by spaces"
line <- getLine
let numberStrings = words line
let numbers = map read numberStrings
putStrLn $ "The result of the calculation is " ++ (show $ foldr1 (*) numbers + 10)
Lots of code inside IO here. Let's extract some functions:
main :: IO ()
main = do
putStrLn "Enter the numbers you want me to process, separated by spaces"
result <- fmap processLine getLine -- fmap :: (a -> b) -> IO a -> IO b
-- runs an impure result through a pure function
-- without leaving IO
putStrLn $ "The result of the calculation is " ++ result
processLine :: String -> String -- look ma, no IO!
processLine = show . calculate . readNumbers
readNumbers :: String -> [Int]
readNumbers = map read . words
calculate :: [Int] -> Int
calculate numbers = product numbers + 10
product :: [Int] -> Int
product = foldr1 (*)
I've pulled logic out of main into pure functions which are easier to read, easier for the compiler to optimise, and more reusable (and so more testable). The program as a whole still lives inside IO because the data is obtained impurely (see the last part of this answer for a more thorough treatment of this argument). Impure data can be piped through pure functions using fmap and other combinators; you should try to put as little logic in main as possible.
Your code does seem to be most of the way there; as others have suggested you could extract lines 2-4 of your main into another function.
In other words, how can I implement (without getting all sorts of errors about IO stuff) the function whose type would be
getBarsFromFile :: Filename -> OHLCBarList
so that the grunt work that was being done in the first four lines of main can be properly encapsulated?
You cannot do this without getting all sorts of errors about IO stuff because this type for getBarsFromFile misses an IO. Probably that's what the errors about IO stuff are trying to tell you. Did you try understanding and fixing the errors?
In your situation, I would start by abstracting over the second to fourth line of your main in a function:
parseBars :: ByteString -> OHLCBarList
And then I would combine this function with getContentsOfFile to get:
getBarsFromFile :: FilePath -> IO OHLCBarList
This I would call in main.
I' ve got a problem with Haskell. I have text file looking like this:
5.
7.
[(1,2,3),(4,5,6),(7,8,9),(10,11,12)].
I haven't any idea how can I get the first 2 numbers (2 and 7 above) and the list from the last line. There are dots on the end of each line.
I tried to build a parser, but function called 'readFile' return the Monad called IO String. I don't know how can I get information from that type of string.
I prefer work on a array of chars. Maybe there is a function which can convert from 'IO String' to [Char]?
I think you have a fundamental misunderstanding about IO in Haskell. Particularly, you say this:
Maybe there is a function which can convert from 'IO String' to [Char]?
No, there isn't1, and the fact that there is no such function is one of the most important things about Haskell.
Haskell is a very principled language. It tries to maintain a distinction between "pure" functions (which don't have any side-effects, and always return the same result when give the same input) and "impure" functions (which have side effects like reading from files, printing to the screen, writing to disk etc). The rules are:
You can use a pure function anywhere (in other pure functions, or in impure functions)
You can only use impure functions inside other impure functions.
The way that code is marked as pure or impure is using the type system. When you see a function signature like
digitToInt :: String -> Int
you know that this function is pure. If you give it a String it will return an Int and moreover it will always return the same Int if you give it the same String. On the other hand, a function signature like
getLine :: IO String
is impure, because the return type of String is marked with IO. Obviously getLine (which reads a line of user input) will not always return the same String, because it depends on what the user types in. You can't use this function in pure code, because adding even the smallest bit of impurity will pollute the pure code. Once you go IO you can never go back.
You can think of IO as a wrapper. When you see a particular type, for example, x :: IO String, you should interpret that to mean "x is an action that, when performed, does some arbitrary I/O and then returns something of type String" (note that in Haskell, String and [Char] are exactly the same thing).
So how do you ever get access to the values from an IO action? Fortunately, the type of the function main is IO () (it's an action that does some I/O and returns (), which is the same as returning nothing). So you can always use your IO functions inside main. When you execute a Haskell program, what you are doing is running the main function, which causes all the I/O in the program definition to actually be executed - for example, you can read and write from files, ask the user for input, write to stdout etc etc.
You can think of structuring a Haskell program like this:
All code that does I/O gets the IO tag (basically, you put it in a do block)
Code that doesn't need to perform I/O doesn't need to be in a do block - these are the "pure" functions.
Your main function sequences together the I/O actions you've defined in an order that makes the program do what you want it to do (interspersed with the pure functions wherever you like).
When you run main, you cause all of those I/O actions to be executed.
So, given all that, how do you write your program? Well, the function
readFile :: FilePath -> IO String
reads a file as a String. So we can use that to get the contents of the file. The function
lines:: String -> [String]
splits a String on newlines, so now you have a list of Strings, each corresponding to one line of the file. The function
init :: [a] -> [a]
Drops the last element from a list (this will get rid of the final . on each line). The function
read :: (Read a) => String -> a
takes a String and turns it into an arbitrary Haskell data type, such as Int or Bool. Combining these functions sensibly will give you your program.
Note that the only time you actually need to do any I/O is when you are reading the file. Therefore that is the only part of the program that needs to use the IO tag. The rest of the program can be written "purely".
It sounds like what you need is the article The IO Monad For People Who Simply Don't Care, which should explain a lot of your questions. Don't be scared by the term "monad" - you don't need to understand what a monad is to write Haskell programs (notice that this paragraph is the only one in my answer that uses the word "monad", although admittedly I have used it four times now...)
Here's the program that (I think) you want to write
run :: IO (Int, Int, [(Int,Int,Int)])
run = do
contents <- readFile "text.txt" -- use '<-' here so that 'contents' is a String
let [a,b,c] = lines contents -- split on newlines
let firstLine = read (init a) -- 'init' drops the trailing period
let secondLine = read (init b)
let thirdLine = read (init c) -- this reads a list of Int-tuples
return (firstLine, secondLine, thirdLine)
To answer npfedwards comment about applying lines to the output of readFile text.txt, you need to realize that readFile text.txt gives you an IO String, and it's only when you bind it to a variable (using contents <-) that you get access to the underlying String, so that you can apply lines to it.
Remember: once you go IO, you never go back.
1 I am deliberately ignoring unsafePerformIO because, as implied by the name, it is very unsafe! Don't ever use it unless you really know what you are doing.
As a programming noob, I too was confused by IOs. Just remember that if you go IO you never come out. Chris wrote a great explanation on why. I just thought it might help to give some examples on how to use IO String in a monad. I'll use getLine which reads user input and returns an IO String.
line <- getLine
All this does is bind the user input from getLine to a value named line. If you type this this in ghci, and type :type line it will return:
:type line
line :: String
But wait! getLine returns an IO String
:type getLine
getLine :: IO String
So what happened to the IOness from getLine? <- is what happened. <- is your IO friend. It allows you to bring out the value that is tainted by the IO within a monad and use it with your normal functions. Monads are easily identified because they begin with do. Like so:
main = do
putStrLn "How much do you love Haskell?"
amount <- getLine
putStrln ("You love Haskell this much: " ++ amount)
If you're like me, you'll soon discover that liftIO is your next best monad friend, and that $ help reduce the number of parenthesis you need to write.
So how do you get the information from readFile? Well if readFile's output is IO String like so:
:type readFile
readFile :: FilePath -> IO String
Then all you need is your friendly <-:
yourdata <- readFile "samplefile.txt"
Now if type that in ghci and check the type of yourdata you'll notice it's a simple String.
:type yourdata
text :: String
As people already say, if you have two functions, one is readStringFromFile :: FilePath -> IO String, and another is doTheRightThingWithString :: String -> Something, then you don't really need to escape a string from IO, since you can combine this two functions in various ways:
With fmap for IO (IO is Functor):
fmap doTheRightThingWithString readStringFromFile
With (<$>) for IO (IO is Applicative and (<$>) == fmap):
import Control.Applicative
...
doTheRightThingWithString <$> readStringFromFile
With liftM for IO (liftM == fmap):
import Control.Monad
...
liftM doTheRightThingWithString readStringFromFile
With (>>=) for IO (IO is Monad, fmap == (<$>) == liftM == \f m -> m >>= return . f):
readStringFromFile >>= \string -> return (doTheRightThingWithString string)
readStringFromFile >>= \string -> return $ doTheRightThingWithString string
readStringFromFile >>= return . doTheRightThingWithString
return . doTheRightThingWithString =<< readStringFromFile
With do notation:
do
...
string <- readStringFromFile
-- ^ you escape String from IO but only inside this do-block
let result = doTheRightThingWithString string
...
return result
Every time you will get IO Something.
Why you would want to do it like that? Well, with this you will have pure and
referentially transparent programs (functions) in your language. This means that every function which type is IO-free is pure and referentially transparent, so that for the same arguments it will returns the same values. For example, doTheRightThingWithString would return the same Something for the same String. However readStringFromFile which is not IO-free can return different strings every time (because file can change), so that you can't escape such unpure value from IO.
If you have a parser of this type:
myParser :: String -> Foo
and you read the file using
readFile "thisfile.txt"
then you can read and parse the file using
fmap myParser (readFile "thisfile.txt")
The result of that will have type IO Foo.
The fmap means myParser runs "inside" the IO.
Another way to think of it is that whereas myParser :: String -> Foo, fmap myParser :: IO String -> IO Foo.
I want to write functions and put result to string.
I want function:
read' :: FilePath -> String
I use:
:t readFile
readFile :: FilePath -> IO String
I make:
read' :: IO ()
read' = do
str <- readFile "/home/shk/workspace/src/test.txt"
putStrLn str
I want to ask str is string or not?
We know that:
:t putStrLn
putStrLn :: String -> IO ()
Then why i can't:
read' :: String
read' = do
str <- readFile "/home/shk/workspace/lxmpp/src/test.txt"
str
I get error that:
Couldn't match expected type `[t0]' with actual type `IO String'
In the return type of a call of `readFile'
In a stmt of a 'do' expression:
str <- readFile "/home/shk/workspace/lxmpp/src/test.txt"
In the expression:
do { str <- readFile "/home/shk/workspace/src/test.txt";
str }
Thank you.
Just to quibble a bit more, while the other answers are perfectly correct, I want to emphasize something: Something with the type IO String isn't just a string that the type system won't let you get at directly. It's a computation that performs I/O to get a string for you. Applying readFile to a file path doesn't return a String value any more than putting a steak next to a meat grinder magically turns them into a hamburger.
When you have some code like this:
foo = do let getStr = readFile "input.txt"
s1 <- getStr
s2 <- getStr
-- etc.
That doesn't mean you're "taking the string out of getStr twice". It means you're performing the computation twice and can easily get different results between the two.
I think no one has answered this, very important question, yet:
I want to ask str is string or not?
I will try to.
The type of the variable str is String, yes.
However, the scope of this variable is very limited. I think desugaring the do-notation is necessary for understanding:
read' = readFile "/home/shk/workspace/src/test.txt" >>= (\str -> putStrLn str)
I think here it becomes more clear why str is not good enough. It is an argument of the function you pass to >>=. Its value only becomes available when someone calls your function, which happens only when the IO action containing it is being executed.
Also, the type of read' :: IO () is determined not so much by the putStrLn str, but rather by the return type of the operator >>=. Have a look at it (specialized to the IO monad):
(>>=) :: IO a -> (a -> IO b) -> IO b
You can see that the result is always an IO b action, so trying to change any of arguments won't help.
You can read some monad tutorial if you want to understand why the type is the way it is. The intuition behind it is: you can't perform an action without performing an action.
And on the practical side of the question, to use the value returned by some action, instead of trying to do use (extractValue inputAction), which does not make sense because extractValue is not possible, try inputAction >>= use if your use does involve I/O, or fmap use inputAction if it does not.
You should use return str in read' if you want it to return str instead of (). You can't strip IO from the type of read', since it's not a pure function. To get a better grip on how input/output in Haskell works I recommend you to read a tutorial.
As a more detailed reason why: It allows impurity.
You absolutely can not perform IO during a pure operation, or it would completely break referential transparency. Technically you can use unsafePerformIO but it would break referential transparency in this case - you should only use that if you can guarantee that the result is always the same.