I am beginner in programming via Haskell language. I am interested at writing function in Haskell, which has side effects like this (for educational purposes, it can have another side effect):
-- function foo behaves like random numbers generator
foo :: Int
{- foo = some magic -}
-- It can print True or False
print $ show (foo /= foo)
As I understood, I have to implement some type instance of Monad with such behaviour (like this: IO Why are side-effects modeled as monads in Haskell?). But I really don't understand how to write simple example of function for this purpose.
If foo really behaves like some RNG, it must have some monadic type, or take as an additional input the state of the RNG.
Here's an example, using the system RNG (which lives in the IO monad):
import System.Random
foo :: IO Int
foo = randomRIO (1,5) -- random number between 1 and 5
main :: IO ()
main = do
x <- foo
y <- foo
print $ x /= y
In main, we simply call foo twice, binding its results to x,y, which have type Int. Then, we can compare them as usual.
Note that we can not use foo /= foo since foo :: IO Int, an action returning an Int, and we can't compare actions, only basic data like integers. This is why we first execute the actions, and then compare their results.
The main could be written in one line exploiting some more advanced combinators
main = print =<< ((/=) <$> foo <*> foo)
-- or
main = print =<< liftA2 (/=) foo foo
but I would not focus on these at the beginning.
A more general example of a side-effecful computation:
foo :: IORef String -> IO Int
foo s = do
x <- readIORef s -- reading a "variable"
print x -- output
w <- getLine -- input
writeIORef s ("qwerty" ++ x) -- writing a variable
length <$> readIORef s
main :: IO ()
main = do
s <- newIORef "dummy"
print =<< foo s
Related
I am trying to get a good grip on the do notation in Haskell.
I could use it with Maybe and then print the result. Like this:
maybeAdd :: Maybe Integer
maybeAdd = do one <- maybe1
two <- maybe2
three <- maybe3
return (one + two + three)
main :: IO ()
main = putStr (show $ fromMaybe 0 maybeAdd)
But instead of having a separate function I am trying to use the do notation with the Maybe inside the main function. But I am not having any luck. The various attempts I tried include:
main :: IO ()
main = do one <- maybe1
two <- maybe2
three <- maybe3
putStr (show $ fromMaybe 0 $ return (one + two + three))
main :: IO ()
main = do one <- maybe1
two <- maybe2
three <- maybe3
putStr (show $ fromMaybe 0 $ Just (one + two + three))
main :: IO ()
main = do one <- maybe1
two <- maybe2
three <- maybe3
putStr (show $ (one + two + three))
All of these leads to various types of compilation errors, which unfortunately I failed to decipher to get the correct way to do it.
How do I achieve the above? And perhaps maybe an explanation of why the approaches I tried were wrong also?
Each do block must work within a single monad. If you want to use multiple monads, you could use multiple do blocks. Trying to adapt your code:
main :: IO ()
main = do -- IO block
let x = do -- Maybe block
one <- maybe1
two <- maybe2
three <- maybe3
return (one + two + three)
putStr (show $ fromMaybe 0 x)
You could even use
main = do -- IO block
putStr $ show $ fromMaybe 0 $ do -- Maybe block
one <- maybe1
two <- maybe2
three <- maybe3
return (one + two + three)
-- other IO actions here
but it could be less readable in certain cases.
The MaybeT monad transformer would come handy in this particular case. MaybeT monad transformer is just a type defined something like;
newtype MaybeT m a = MaybeT {runMaybeT :: m (Maybe a)}
Actually transformers like MaybeT, StateT etc, are readily available in Control.Monad.Trans.Maybe, Control.Monad.Trans.State... For illustration purposes it' Monad instance could be something like shown below;
instance Monad m => Monad (MaybeT m) where
return = MaybeT . return . Just
x >>= f = MaybeT $ runMaybeT x >>= g
where
g Nothing = return Nothing
g (Just x) = runMaybeT $ f x
so as you will notice the monadic f function takes a value that resides in the Maybe monad which itself is in another monad (IO in our case). The f function does it's thing and wraps the result back into MaybeT m a.
Also there is a MonadTrans class where you can have some common functionalities those are used by the transformer types. One such is lift which is used to lift the value into a transformer according to that particular instance's definition. For MaybeT it should look like
instance MonadTrans MaybeT where
lift = MaybeT . (liftM Just)
Lets perform your task with monad transformers.
addInts :: MaybeT IO ()
addInts = do
lift $ putStrLn "Enter two integers.."
i <- lift getLine
guard $ test i
j <- lift getLine
guard $ test j
lift . print $ (read i :: Int) + (read j :: Int)
where
test = and . (map isDigit)
So when called like
λ> runMaybeT addInts
Enter two integers..
1453
1571
3024
Just ()
The catch is, since a monad transformer is also a member of Monad typeclass, one can nest them indefinitelly and still do things under a singe do notation.
Edit: answer gets downvoted but it is unclear to me why. If there is something wrong with the approach please care to elaborate me so that it helps people including me to learn something better.
Taking the opportunity of being on the edit session, i would like to add a better code since i think Char based testing might not be the best idea as it will not take negative Ints into account. So let's try using readMaybe from the Text.Read package while we are doing things with the Maybe type.
import Control.Monad.Trans.Maybe
import Control.Monad.Trans.Class (lift)
import Text.Read (readMaybe)
addInts :: MaybeT IO ()
addInts = do
lift $ putStrLn "Enter two integers.."
i <- lift getLine
MaybeT $ return (readMaybe i :: Maybe Int)
j <- lift getLine
MaybeT $ return (readMaybe j :: Maybe Int)
lift . print $ (read i :: Int) + (read j :: Int)
I guess now it works better...
λ> runMaybeT addInts
Enter two integers..
-400
500
100
Just ()
λ> runMaybeT addInts
Enter two integers..
Not an Integer
Nothing
I'm fairly new to Haskell and have been trying to find a way to pass multiple IO-tainted values to a function to deal with a C library. Most people seem to use the <- operator inside a do block, like this:
g x y = x ++ y
interactiveConcat1 = do {x <- getLine;
y <- getLine;
putStrLn (g x y);
return ()}
This makes me feel like I'm doing C, except emacs can't auto-indent. I tried to write this in a more Lispy style:
interactiveConcat2 = getLine >>= (\x ->
getLine >>= (\y ->
putStrLn (g x y) >>
return () ))
That looks like a mess, and has a string of closed parentheses you have to count at the end (except again, emacs can reliably assist with this task in Lisp, but not in Haskell). Yet another way is to say
import Control.Applicative
interactiveConcat3 = return g <*> getLine <*> getLine >>= putStrLn
which looks pretty neat but isn't part of the base language.
Is there any less laborious notation for peeling values out of the IO taint boxes? Perhaps there is a cleaner way using a lift* or fmap? I hope it isn't too subjective to ask what is considered "idiomatic"?
Also, any tips for making emacs cooperate better than (Haskell Ind) mode would be greatly appreciated. Thanks!
John
Edit: I stumbled across https://wiki.haskell.org/Do_notation_considered_harmful and realized that the nested parentheses in the lambda chain I wrote is not necessary. However it seems the community (and ghc implementors) have embraced the Applicative-inspired style using , <*>, etc, which seems to make the code easier to read in spite of the headaches with figuring out operator precedence.
Note: This post is written in literate Haskell. You can save it as Main.lhs and try it in your GHCi.
A short remark first: you can get rid of the semicolons and the braces in do. Also, putStrLn has type IO (), so you don't need return ():
interactiveConcat1 = do
x <- getLine
y <- getLine
putStrLn $ g x y
We're going to work with IO, so importing Control.Applicative or Control.Monad will come in handy:
> module Main where
> import Control.Applicative
> -- Repeat your definition for completeness
> g :: [a] -> [a] -> [a]
> g = (++)
You're looking for something like this:
> interactiveConcat :: IO ()
> interactiveConcat = magic g getLine getLine >>= putStrLn
What type does magic need? It returns a IO String, takes a function that returns an String and takes usual Strings, and takes two IO Strings:
magic :: (String -> String -> String) -> IO String -> IO String -> IO String
We can probably generalize this type to
> magic :: (a -> b -> c) -> IO a -> IO b -> IO c
A quick hoogle search reveals that there are already two functions with almost that type: liftA2 from Control.Applicative and liftM2 from Control.Monad. They're defined for every Applicative and – in case of liftM2 – Monad. Since IO is an instance of both, you can choose either one:
> magic = liftA2
If you use GHC 7.10 or higher, you can also use <$> and <*> without import and write interactiveConcat as
interactiveConcat = g <$> getLine <*> getLine >>= putStrLn
For completeness, lets add a main so that we can easily check this functionality via runhaskell Main.lhs:
> main :: IO ()
> main = interactiveConcat
A simple check shows that it works as intended:
$ echo "Hello\nWorld" | runhaskell Main.lhs
HelloWorld
References
Applicative in the Typeclassopedia
The section "Some useful monadic functions" of LYAH's chapter "For a Few Monads More".
You can use liftA2 (or liftM2 from Control.Monad):
import Control.Applicative (liftA2)
liftA2 g getLine getLine >>= putStrLn
I'm trying to work out if it's possible to write an abstraction for the following situation. Suppose I have a type a with function a -> m Bool e.g. MVar Bool and readMVar. To abstract this concept out I create a newtype wrapper for the type and its function:
newtype MPredicate m a = MPredicate (a,a -> m Bool)
I can define a fairly simple operation like so:
doUnless :: (Monad m) => Predicate m a -> m () -> m ()
doUnless (MPredicate (a,mg)) g = mg a >>= \b -> unless b g
main = do
b <- newMVar False
let mpred = MPredicate (b,readMVar)
doUnless mpred (print "foo")
In this case doUnless would print "foo". Aside: I'm not sure whether a type class might be more appropriate to use instead of a newtype.
Now take the code below, which outputs an incrementing number then waits a second and repeats. It does this until it receives a "turn off" instruction via the MVar.
foobar :: MVar Bool -> IO ()
foobar mvb = foobar' 0
where
foobar' :: Int -> IO ()
foobar' x = readMVar mvb >>= \b -> unless b $ do
let x' = x + 1
print x'
threadDelay 1000000
foobar' x'
goTillEnter :: MVar Bool -> IO ()
goTillEnter mv = do
_ <- getLine
_ <- takeMVar mv
putMVar mv True
main = do
mvb <- newMVar False
forkIO $ foobar mvb
goTillEnter mvb
Is it possible to refactor foobar so that it uses MPredicate and doUnless?
Ignoring the actual implementation of foobar' I can think of a simplistic way of doing something similar:
cycleUnless :: x -> (x -> x) -> MPredicate m a -> m ()
cycleUnless x g mp = let g' x' = doUnless mp (g' $ g x')
in g' $ g x
Aside: I feel like fix could be used to make the above neater, though I still have trouble working out how to use it
But cycleUnless won't work on foobar because the type of foobar' is actually Int -> IO () (from the use of print x').
I'd also like to take this abstraction further, so that it can work threading around a Monad. With stateful Monads it becomes even harder. E.g.
-- EDIT: Updated the below to show an example of how the code is used
{- ^^ some parent function which has the MVar ^^ -}
cycleST :: (forall s. ST s (STArray s Int Int)) -> IO ()
cycleST sta = readMVar mvb >>= \b -> unless b $ do
n <- readMVar someMVar
i <- readMVar someOtherMVar
let sta' = do
arr <- sta
x <- readArray arr n
writeArray arr n (x + i)
return arr
y = runSTArray sta'
print y
cycleST sta'
I have something similar to the above working with RankNTypes. Now there's the additional problem of trying to thread through the existential s, which is not likely to type check if threaded around through an abstraction the likes of cycleUnless.
Additionally, this is simplified to make the question easier to answer. I also use a set of semaphores built from MVar [MVar ()] similar to the skip channel example in the MVar module. If I can solve the above problem I plan to generalize the semaphores as well.
Ultimately this isn't some blocking problem. I have 3 components of the application operating in a cycle off the same MVar Bool but doing fairly different asynchronous tasks. In each one I have written a custom function that performs the appropriate cycle.
I'm trying to learn the "don't write large programs" approach. What I'd like to do is refactor chunks of code into their own mini libraries so that I'm not building a large program but assembling lots of small ones. But so far this particular abstraction is escaping me.
Any thoughts on how I might go about this are very much appreciated!
You want to cleanly combine a stateful action having side effects, a delay, and an independent stopping condition.
The iterative monad transformer from the free package can be useful in these cases.
This monad transformer lets you describe a (possibly nonending) computation as a series of discrete steps. And what's better, it let's you interleave "stepped" computations using mplus. The combined computation stops when any of the individual computations stops.
Some preliminary imports:
import Data.Bool
import Control.Monad
import Control.Monad.Trans
import Control.Monad.Trans.Iter (delay,untilJust,IterT,retract,cutoff)
import Control.Concurrent
Your foobar function could be understood as a "sum" of three things:
A computation that does nothing but reading from the MVar at each step, and finishes when the Mvar is True.
untilTrue :: (MonadIO m) => MVar Bool -> IterT m ()
untilTrue = untilJust . liftM guard . liftIO . readMVar
An infinite computation that takes a delay at each step.
delays :: (MonadIO m) => Int -> IterT m a
delays = forever . delay . liftIO . threadDelay
An infinite computation that prints an increasing series of numbers.
foobar' :: (MonadIO m) => Int -> IterT m a
foobar' x = do
let x' = x + 1
liftIO (print x')
delay (foobar' x')
With this in place, we can write foobar as:
foobar :: (MonadIO m) => MVar Bool -> m ()
foobar v = retract (delays 1000000 `mplus` untilTrue v `mplus` foobar' 0)
The neat thing about this is that you can change or remove the "stopping condition" and the delay very easily.
Some clarifications:
The delay function is not a delay in IO, it just tells the iterative monad transformer to "put the argument in a separate step".
retract brings you back from the iterative monad transformer to the base monad. It's like saying "I don't care about the steps, just run the computation". You can combine retract with cutoff if you want to limit the maximum number of iterations.
untilJustconverts a value m (Maybe a) of the base monad into a IterT m a by retrying in each step until a Just is returned. Of course, this risks non-termination!
MPredicate is rather superfluous here; m Bool can be used instead. The monad-loops package contains plenty of control structures with m Bool conditions. whileM_ in particular is applicable here, although we need to include a State monad for the Int that we're threading around:
import Control.Monad.State
import Control.Monad.Loops
import Control.Applicative
foobar :: MVar Bool -> IO ()
foobar mvb = (`evalStateT` (0 :: Int)) $
whileM_ (not <$> lift (readMVar mvb)) $ do
modify (+1)
lift . print =<< get
lift $ threadDelay 1000000
Alternatively, we can use a monadic version of unless. For some reason monad-loops doesn't export such a function, so let's write it:
unlessM :: Monad m => m Bool -> m () -> m ()
unlessM mb action = do
b <- mb
unless b action
It's somewhat more convenient and more modular in a monadic setting, since we can always go from a pure Bool to m Bool, but not vice versa.
foobar :: MVar Bool -> IO ()
foobar mvb = go 0
where
go :: Int -> IO ()
go x = unlessM (readMVar mvb) $ do
let x' = x + 1
print x'
threadDelay 1000000
go x'
You mentioned fix; sometimes people indeed use it for ad-hoc monadic loops, for example:
printUntil0 :: IO ()
printUntil0 =
putStrLn "hello"
fix $ \loop -> do
n <- fmap read getLine :: IO Int
print n
when (n /= 0) loop
putStrLn "bye"
With some juggling it's possible to use fix with multi-argument functions. In the case of foobar:
foobar :: MVar Bool -> IO ()
foobar mvb = ($(0 :: Int)) $ fix $ \loop x -> do
unlessM (readMVar mvb) $ do
let x' = x + 1
print x'
threadDelay 1000000
loop x'
I'm not sure what's your MPredicate is doing.
First, instead of newtyping a tuple, it's probably better to use a normal algebric data type
data MPredicate a m = MPredicate a (a -> m Bool)
Second, the way you use it, MPredicate is equivalent to m Bool.
Haskell is lazzy, therefore there is no need to pass, a function and it's argument (even though
it's usefull with strict languages). Just pass the result, and the function will be called when needed.
I mean, instead of passing (x, f) around, just pass f x
Of course, if you are not trying to delay the evaluation and really need at some point, the argument or the function as well as the result, a tuple is fine.
Anyway, in the case your MPredicate is only there to delay the function evaluation, MPredicat reduces to m Bool and doUnless to unless.
Your first example is strictly equivalent :
main = do
b <- newMVar False
unless (readMVar b) (print "foo")
Now, if you want to loop a monad until a condition is reach (or equivalent) you should have a look at the monad-loop package. What you are looking it at is probably untilM_ or equivalent.
Being quite new to Haskell, I'm currently trying to improve my skills by writing an interpreter for a simple imperative toy language.
One of the expressions in this language is input, which reads a single integer from standard input. However, when I assign the value of this expression to a variable and then use this variable later, it seems ot me that I actually stored the computation of reading a value rather the read value itself. This means that e.g. the statements
x = input;
y = x + x;
will cause the interpreter to invoke the input procedure three times rather than one.
Internally in the evaluator module, I use a Map to store the values of variables. Because I need to deal with IO, this gets wrapped in an IO monad, as immortalized in the following minimal example:
import qualified Data.Map as Map
type State = Map.Map String Int
type Op = Int -> Int -> Int
input :: String -> IO State -> IO State
input x state = do line <- getLine
st <- state
return $ Map.insert x (read line) st
get :: String -> IO State -> IO Int
get x state = do st <- state
return $ case Map.lookup x st of
Just i -> i
eval :: String -> Op -> String -> IO State -> IO Int
eval l op r state = do i <- get l state
j <- get r state
return $ op i j
main :: IO ()
main = do let state = return Map.empty
let state' = input "x" state
val <- eval "x" (+) "x" state'
putStrLn . show $ val
The second line in the main function simulates the assignment of x, while the third line simulates the evaluation of the binary + operator.
My question is: How do I get around this, such that the code above only inputs once? I suspect that it is the IO-wrapping that causes the problem, but as we're dealing with IO I see no way out of that..?
Remember that IO State is not an actual state, but instead the specification for an IO machine which eventually produces a State. Let's consider input as an IO-machine transformer
input :: String -> IO State -> IO State
input x state = do line <- getLine
st <- state
return $ Map.insert x (read line) st
Here, provided a machine for producing a state, we create a bigger machine which takes that passed state and adding a read from an input line. Again, to be clear, input name st is an IO-machine which is a slight modification of the IO-machine st.
Let's now examine get
get :: String -> IO State -> IO Int
get x state = do st <- state
return $ case Map.lookup x st of
Just i -> i
Here we have another IO-machine transformer. Given a name and an IO-machine which produces a State, get will produce an IO-machine which returns a number. Note again that get name st is fixed to always use the state produced by the (fixed, input) IO-machine st.
Let's combine these pieces in eval
eval :: String -> Op -> String -> IO State -> IO Int
eval l op r state = do i <- get l state
j <- get r state
return $ op i j
Here we call get l and get r each on the same IO-machine state and thus produce two (completely independent) IO-machines get l state and get r state. We then evaluate their IO effects one after another and return the op-combination of their results.
Let's examine the kinds of IO-machines built in main. In the first line we produce a trivial IO-machine, called state, written return Map.empty. This IO-machine, each time it's run, performs no side effects in order to return a fresh, blank Map.Map.
In the second line, we produce a new kind of IO-machine called state'. This IO-machine is based off of the state IO-machine, but it also requests an input line. Thus, to be clear, each time state' runs, a fresh Map.Map is generated and then an input line is read to read some Int, stored at "x".
It should be clear where this is going, but now when we examine the third line we see that we pass state', the IO-machine, into eval. Previously we stated that eval runs its input IO-machine twice, once for each name, and then combines the results. By this point it should be clear what's happening.
All together, we build a certain kind of machine which draws input and reads it as an integer, assigning it to a name in a blank Map.Map. We then build this IO-machine into a larger one which uses the first IO-machine twice, in two separate invocations, in order to collect data and combine it with an Op.
Finally, we run this eval machine using do notation (the (<-) arrow indicates running the machine). Clearly it should collect two separate lines.
So what do we really want to do? Well, we need to simulate ambient state in the IO monad, not just pass around Map.Maps. This is easy to do by using an IORef.
import Data.IORef
input :: IORef State -> String -> IO ()
input ref name = do
line <- getLine
modifyIORef ref (Map.insert name (read line))
eval :: IORef State -> Op -> String -> String -> IO Int
eval ref op l r = do
stateSnapshot <- readIORef ref
let Just i = Map.lookup l stateSnapshot
Just j = Map.lookup l stateSnapshot
return (op i j)
main = do
st <- newIORef Map.empty -- create a blank state, embedded into IO, not a value
input st "x" -- request input *once*
val <- eval st (+) "x" "x" -- compute the op
putStrLn . show $ val
It's fine to wrap your actions such as getLine in IO, but to me it looks like your problem is that you're trying to pass your state in the IO monad. Instead, I think this is probably time you get introduced to monad transformers and how they'll let you layer the IO and State monads to get the functionality of both in one.
Monad transformers are a pretty complex topic and it'll take a while to get to where you're comfortable with them (I'm still learning new things all the time about them), but they're a very useful tool when you need to layer multiple monads. You'll need the mtl library to follow this example.
First, imports
import qualified Data.Map as Map
import Control.Monad.State
Then types
type Op = Int -> Int -> Int
-- Renamed to not conflict with Control.Monad.State.State
type AppState = Map.Map String Int
type Interpreter a = StateT AppState IO a
Here Interpreter is the Monad in which we'll build our interpreter. We also need a way to run the interpreter
-- A utility function for kicking off an interpreter
runInterpreter :: Interpreter a -> IO a
runInterpreter interp = evalStateT interp Map.empty
I figured defaulting to Map.empty was sufficient.
Now, we can build our interpreter actions in our new monad. First we start with input. Instead of returning our new state, we just modify what is current in our map:
input :: String -> Interpreter ()
input x = do
-- IO actions have to be passed to liftIO
line <- liftIO getLine
-- modify is a member of the MonadState typeclass, which StateT implements
modify (Map.insert x (read line))
I had to rename get so that it didn't conflict with get from Control.Monad.State, but it does basically the same thing as before, it just takes our map and looks up that variable in it.
-- Had to rename to not conflict with Control.Monad.State.get
-- Also returns Maybe Int because it's safer
getVar :: String -> Interpreter (Maybe Int)
getVar x = do
-- get is a member of MonadState
vars <- get
return $ Map.lookup x vars
-- or
-- get x = fmap (Map.lookup x) get
Next, eval now just looks up each variable in our map, then uses liftM2 to keep the return value as Maybe Int. I prefer the safety of Maybe, but you can rewrite it if you prefer
eval :: String -> Op -> String -> Interpreter (Maybe Int)
eval l op r = do
i <- getVar l
j <- getVar r
-- liftM2 op :: Maybe Int -> Maybe Int -> Maybe Int
return $ liftM2 op i j
Finally, we write our sample program. It stores user input to the variable "x", adds it to itself, and prints out the result.
-- Now we can write our actions in our own monad
program :: Interpreter ()
program = do
input "x"
y <- eval "x" (+) "x"
case y of
Just y' -> liftIO $ putStrLn $ "y = " ++ show y'
Nothing -> liftIO $ putStrLn "Error!"
-- main is kept very simple
main :: IO ()
main = runInterpreter program
The basic idea is that there is a "base" monad, here IO, and these actions are "lifted" up to the "parent" monad, here StateT AppState. There is a typeclass implementation for the different state operations get, put, and modify in the MonadState typeclass, which StateT implements, and in order to lift IO actions there's a pre-made liftIO function that "lifts" IO actions to the parent monad. Now we don't have to worry about passing around our state explicitly, we can still perform IO, and it has even simplified the code!
I would recommend reading the Real World Haskell chapter on monad transformers to get a better feel for them. There are other useful ones as well, such as ErrorT for handling errors, ReaderT for static configuration, WriterT for aggregating results (usually used for logging), and many others. These can be layered into what is called a transformer stack, and it's not too difficult to make your own either.
Instead of passing an IO State, you can pass State and then use higher-level functions to deal with IO. You can go further and make get and eval free from side-effects:
input :: String -> State -> IO State
input x state = do
line <- getLine
return $ Map.insert x (read line) state
get :: String -> State -> Int
get x state = case Map.lookup x state of
Just i -> i
eval :: String -> Op -> String -> State -> Int
eval l op r state = let i = get l state
j = get r state
in op i j
main :: IO ()
main = do
let state = Map.empty
state' <- input "x" state
let val = eval "x" (+) "x" state'
putStrLn . show $ val
If you're actually building an interpreter, you'll presumably have a list of instructions to execute at some point.
This is my rough translation of your code (although I'm only a beginner myself)
import Data.Map (Map, empty, insert, (!))
import Control.Monad (foldM)
type ValMap = Map String Int
instrRead :: String -> ValMap -> IO ValMap
instrRead varname mem = do
putStr "Enter an int: "
line <- getLine
let intval = (read line)::Int
return $ insert varname intval mem
instrAdd :: String -> String -> String -> ValMap -> IO ValMap
instrAdd varname l r mem = do
return $ insert varname result mem
where result = (mem ! l) + (mem ! r)
apply :: ValMap -> (ValMap -> IO ValMap) -> IO ValMap
apply mem instr = instr mem
main = do
let mem0 = empty
let instructions = [ instrRead "x", instrAdd "y" "x" "x" ]
final <- foldM apply mem0 instructions
print (final ! "y")
putStrLn "done"
The foldM applies a function (apply) to a start value (mem0) and a list (instructions) but does so within a monad.
I want to write a function that read some data using getLine and return i.e. a tuple (Integer, Integer) but using do-notation. Something like this (of course it doesn't work):
fun :: (Integer, Integer)
fun = do
a <- read (getLine::Integer)
b <- read (getLine::Integer)
return (a, b)
Do I have to write my own monad for this? Is there any solution to not writing a new monad?
EDIT
So I can write main function that use fun, I think it's the only solution:
main :: IO ()
main = do
tuple <- fun
putStrLn (show tuple)
fun :: IO (Integer, Integer)
fun = do
a1 <- getLine
b1 <- getLine
let a = read (a1)
b = read (b1)
return (a, b)
And above code works.
You type of function should be
fun :: IO (Integer, Integer)
as mentioned by #kaan you should not try to get a mondic value (with side effects) out of the monad as that will break referential transparency. Running fun should always return same value no matter how many times it is run and if we use your type this will not happen. However if the type is IO (Integer, Integer) then it returns the same action every time you use that function and running this action actually perform the side effect of reading the values from the console.
Coming back to using you function. You can do that inside another IO monad like
main = do
(a,b) <- fun
print a
print b
Although there are ways of getting things out of IO using unsafe functions but that is not recommended until you know exactly what you are doing.
As mentioned, you will need to give fun the type IO (Integer, Integer) instead of (Integer, Integer). However, once you have resigned yourself to this fate, there are many ways to skin this cat. Here are a handful of ways to get your imagination going.
fun = do
a <- getLine
b <- getLine
return (read a, read b)
-- import Control.Applicative for (<$>)
-- can also spell (<$>) as fmap, liftA, liftM, and others
fun = do
a <- read <$> getLine
b <- read <$> getLine
return (a, b)
fun = do
a <- readLn
b <- readLn
return (a, b)
fun = liftM2 (,) readLn readLn
-- different type!
-- use in main like this:
-- main = do
-- [a, b] <- fun
-- foo
-- import Control.Monad for replicateM
fun :: IO [Integer]
fun = replicateM 2 readLn