How to avoid the IO monad when solving arithmetic problems in SBV - haskell

I am trying to solve arithmetic problems with SBV.
For example
solution :: SymbolicT IO ()
solution = do
[x, y] <- sFloats ["x", "y"]
constrain $ x + y .<= 2
Main> s1 = sat solution
Main> s2 = isSatisfiable solution
Main> s1
Satisfiable. Model:
x = -1.2030502e-17 :: Float
z = -2.2888208e-37 :: Float
Main> :t s1
s1 :: IO SatResult
Main> s2
True
Main> :t s2
s2 :: IO Bool
While I can do useful things, it is easier for me to work with the pure value (SatResult or Bool) and not with the IO monad.
According to the documentation
sat :: Provable a => a -> IO SatResult
constrain :: SolverContext m => SBool -> m ()
sFloats :: [String] -> Symbolic [SFloat]
type Symbolic = SymbolicT IO
Given the type of functions I use, I understand why I always get to the IO monad.
But looking in the generalized versions of the functions for example sFloats.
sFloats :: MonadSymbolic m => [String] -> m [SFloat]
Depending on type of the function, I can work with a different monad than IO. This gives me hope that we will reach a more useful monad, the Identity monad for example.
Unfortunately looking at the examples always solves the problems within the IO monad, so I couldn't find any examples that would work for me.Besides that I don't have much experience working with monads.
Finally My question is:
Is there any way to avoid the IO monad when solving such a problem with SBV?
Thanks in advance

SBV calls out to the SMT solver of your choice (most likely z3, but others are available too), and presents the results back to you. This means that it performs IO under the hood, and thus you cannot be outside the IO monad. You can create custom monads using MonadSymbolic, but that will not get you out of the IO monad: Since the call to the SMT solver does IO you'll always be in IO.
(And I'd strongly caution against uses of unsafePerformIO as suggested in one of the comments. This is really a bad idea; and you can find lots more information on this elsewhere why you shouldn't do so.)
Note that this is no different than any other IO based computation in Haskell: You perform the IO "in-the-wrapper," but once you get your results, you can do whatever you'd like to do with them in a "pure" environment.
Here's a simple example:
import Data.SBV
import Data.SBV.Control
example :: IO ()
example = runSMT $ do
[x, y] <- sFloats ["x", "y"]
constrain $ x + y .<= 2
query $ do cs <- checkSat
case cs of
Unsat -> io $ putStrLn "Unsatisfiable"
Sat -> do xv <- getValue x
yv <- getValue y
let result = use xv yv
io $ putStrLn $ "Result: " ++ show result
_ -> error $ "Solver said: " ++ show cs
-- Use the results from the solver, in a purely functional way
use :: Float -> Float -> Float
use x y = x + y
Now you can say:
*Main> example
Result: -Infinity
The function example has type IO (), because it does involve calling out to the solver and getting the results. However, once you extract those results (via calls to getValue), you can pass them to the function use which has a very simple purely functional type. So, you keep the "wrapper" in the monad, but actual processing, use-of-the values, etc., remain in the pure world.
Alternatively, you can also extract the values and continue from there:
import Data.SBV
import Data.SBV.Control
example :: IO (Maybe (Float, Float))
example = runSMT $ do
[x, y] <- sFloats ["x", "y"]
constrain $ x + y .<= 2
query $ do cs <- checkSat
case cs of
Unsat -> pure Nothing
Sat -> do xv <- getValue x
yv <- getValue y
pure $ Just (xv, yv)
_ -> error $ "Solver said: " ++ show cs
Now you can say:
*Main> Just (a, b) <- example
*Main> a
-Infinity
*Main> b
4.0302105e-21
Long story short: Don't avoid the IO monad. It's there for a very good reason. Get into it, get your results out, and then the rest of your program can remain purely functional, or whatever other monad you might find yourself in.
Note that none of this is really SBV specific. This is the usual Haskell paradigm of how to use functions with side-effects. (For instance, anytime you use readFile to read the contents of a file to process it further.) Do not try to "get rid of the IO." Instead, simply work with it.

Depending on type of the function, I can work with a different monad than IO.
Not meaningfully different, in the sense you'd hope. Every instance of this class is going to be some transformed version of IO. Sorry!
Time to make a plan that involves understanding and working with IO.

Related

Lazy evaluation of IO actions

I'm trying to write code in source -> transform -> sink style, for example:
let (|>) = flip ($)
repeat 1 |> take 5 |> sum |> print
But would like to do that using IO. I have this impression that my source can be an infinite list of IO actions, and each one gets evaluated once it is needed downstream. Something like this:
-- prints the number of lines entered before "quit" is entered
[getLine..] >>= takeWhile (/= "quit") >>= length >>= print
I think this is possible with the streaming libraries, but can it be done along the lines of what I'm proposing?
Using the repeatM, takeWhile and length_ functions from the streaming library:
import Streaming
import qualified Streaming.Prelude as S
count :: IO ()
count = do r <- S.length_ . S.takeWhile (/= "quit") . S.repeatM $ getLine
print r
This seems to be in that spirit:
let (|>) = flip ($)
let (.>) = flip (.)
getContents >>= lines .> takeWhile (/= "quit") .> length .> print
The issue here is that Monad is not the right abstraction for this, and attempting to do something like this results in a situation where referential transparency is broken.
Firstly, we can do a lazy IO read like so:
module Main where
import System.IO.Unsafe (unsafePerformIO)
import Control.Monad(forM_)
lazyIOSequence :: [IO a] -> IO [a]
lazyIOSequence = pure . go where
go :: [IO a] -> [a]
go (l:ls) = (unsafePerformIO l):(go ls)
main :: IO ()
main = do
l <- lazyIOSequence (repeat getLine)
forM_ l putStrLn
This when run will perform cat. It will read lines and output them. Everything works fine.
But consider changing the main function to this:
main :: IO ()
main = do
l <- lazyIOSequence (map (putStrLn . show) [1..])
putStrLn "Hello World"
This outputs Hello World only, as we didn't need to evaluate any of l. But now consider replacing the last line like the following:
main :: IO ()
main = do
x <- lazyIOSequence (map (putStrLn . show) [1..])
seq (head x) putStrLn "Hello World"
Same program, but the output is now:
1
Hello World
This is bad, we've changed the results of a program just by evaluating a value. This is not supposed to happen in Haskell, when you evaluate something it should just evaluate it, not change the outside world.
So if you restrict your IO actions to something like reading from a file nothing else is reading from, then you might be able to sensibly lazily evaluate things, because when you read from it in relation to all the other IO actions your program is taking doesn't matter. But you don't want to allow this for IO in general, because skipping actions or performing them in a different order can matter (and above, certainly does). Even in the reading a file lazily case, if something else in your program writes to the file, then whether you evaluate that list before or after the write action will affect the output of your program, which again, breaks referential transparency (because evaluation order shouldn't matter).
So for a restricted subset of IO actions, you can sensibly define Functor, Applicative and Monad on a stream type to work in a lazy way, but doing so in the IO Monad in general is a minefield and often just plain incorrect. Instead you want a specialised streaming type, and indeed Conduit defines Functor, Applicative and Monad on a lot of it's types so you can still use all your favourite functions.

Idiomatic Haskell syntax without do-blocks or accumulating parentheses?

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

Write list of random numbers to file. No Instance for (Show (IO a0))

I am trying to write to file a list of random Integers in a file. There seems to be a problem with writeFile here. When I use my function randomFile it says no instance for (Show (IO a0)). I see writeFile doesn't print anything to screen but IO(), so when I call the function randomFile 1 2 3 it says no Instance for Show (IO a0) but actually I just want to execute the function and not have to print anything but how can I avoid this problem. I might be making a lot of errors here. Any help.
import Control.Monad
import Control.Applicative
import System.Random
randNo mind maxd = randomRIO (mind,maxd)
randomFile mind maxd noe = do
let l=(replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd))
writeFile "RFile.txt" (show l)
I think you have a misunderstanding of what IO is. If you haven't done it, I strongly recommend going through the Input and Output section of Learn You a Haskell.
IO doesn't necessarily have anything to do with print. In Haskell every entry in memory that was made by your own code is considered "pure" while any entry that touches the rest of the computer lives in IO (with some exceptions you will learn about over time).
We model IO using something called a Monad. Which you will learn more about the longer you do Haskell. To understand this, let's look at an example of some code that does and doesn't use IO:
noIOused :: Int -> Int
noIOused x = x + 5
usesIO :: Int -> IO Int
usesIO x = print x >> return (x + 5)
usesIO2 :: Int -> IO Int
usesIO2 x = do
print x
return (x + 5)
The first function is "pure". The second and third functions have an IO "effect" that comes in the form of printing to the screen. usesIO and usesIO2 are just 2 different ways of doing the same thing (it's the same code but with different syntax). I'll use the second format, called do notation from here.
Here are some other ways you could have had IO effects:
add5WithFile :: Int -> IO Int
add5WithFile x = do
writeFile "someFile.txt" (show x)
return (x + 5)
Notice that in that function we didn't print anything, we wrote a file. But writing a file has a side effect and interacts with the rest of the system. So any value we return has to get wrapped in IO.
addRandom :: Int -> IO Int
addRandom x = do
y <- randomRIO (1,10)
return (x + y)
In addRandom we called randomRIO (1,10). But the problem is that randomRIO doesn't return an Int. It returns an IO Int. Why? Because in order to get true randomness we need to interact with the system in some way. To get around that, we have to temporarily strip away the IO. That's where this line comes in:
y <- randomRIO (1,10)
That <- arrow tells us that we want a y value outside of IO. For as long as we remain inside the do syntax that y value is going to be "pure". Now we can use it just like any other value.
So for example we couldn't do this:
let w = x + (randomRIO (1,10))
Because that would be trying to add Int to IO Int. And unfortunately our + function doesn't know how to do that. So first we have to "bind" the result of randomRIO to y before we can add it to x.
Now let's look at your code:
let l=(replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd))
writeFile "RFile.txt" (show l)
The type of l is actually IO a0. It's a0 because you haven't told the compiler what kind of number you want. So it doesn't know if you want a fraction, a double, a big integer or whatever.
So the first problem is to let the compiler know a little bit more about what kind of random number you want. We do this by adding a type annotation:
randNo :: Int -> Int -> IO Int
randNo mind maxd = randomRIO (mind,maxd)
Now both you and the compiler knows what kind of value randNo is.
Now we need to "bind" that value inside of the do notation to temporarily escape IO. You might think that would be simple, like this:
randomFile mind maxd noe = do
l <- replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd)
writeFile "RFile.txt" (show l)
Surely that will "bind" the IO Int to l right? Unfortunately not. The problem here is that replicate is a function of the form Int -> a -> [a]. That is, given a number and a type, it will give you a list of that type.
If you give replicate an IO Int it's going to make [IO Int]. That actually looks more like this: List (IO Int) except we use [] as syntactic sugar for lists. Unfortunately if we want to "bind" an IO value to something with <- it has to be the out-most type.
So what you need is a way to turn an [IO Int] into an IO [Int]. There are two ways to do that. If we put \[IO a\] -> IO \[a\] into Hoogle we get this:
sequence :: Monad m => [m a] -> m [a]
As I mentioned before, we generalise IO to something called a Monad. Which isn't really that big a deal, we could pretend that sequence has this signature: sequence :: [IO a] -> IO [a] and it would be the same thing just specialised to IO.
Now your function would be done like this:
randomFile mind maxd noe = do
l <- sequence (replicate (fromInteger(noe ^ noe)) ( mind `randNo` maxd))
writeFile "RFile.txt" (show l)
But a sequence followed by replicate is something people have to do all the time. So someone went and made a function called replicateM:
replicateM :: Monad m => Int -> m a -> m [a]
Now we can write your function like this:
randomFile mind maxd noe = do
l <- replicateM (fromInteger(noe ^ noe)) ( mind `randNo` maxd)
writeFile "RFile.txt" (show l)
And for some real Haskell magic, you can write all 3 lines of code in a single line, like this:
randomFile mind maxd noe = randomRIO >>= writeFile "RFile.txt" . replicateM (fromInteger(noe ^ noe))
If that looks like gibberish to you, then there's a lot you need to learn. Here is the suggested path:
If you haven't already, start from the beginning with Learn You a Haskell
Then learn about how You could have invented Monads
Then learn more about how to use randomness in Haskell
Finally see if you can complete the 20 intermediate Haskell exercises

Abstraction for monadic recursion with "unless"

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.

Haskell monad: IO [Double] to [IO Double]

Consider the following code that is supposed to print out random numbers:
import System.Random.Mersenne
main =
do g <- (newMTGen Nothing)
xs <- (randoms g) :: IO [Double]
mapM_ print xs
When run, I get a segmentation fault error. That is unsurprising, since the function 'randoms' produces an infinite list. Suppose I wanted to print out only the first ten values of xs. How could I do that? xs has type IO [Double], and I think I want a variable of type [IO Double]. What operators exist to convert between the two.
If you get a segmentation fault error, and you didn't use the FFI or any functions with unsafe in their name, that's not unsurprising, in any situation! It means there's a bug with either GHC, or a library you're using is doing something unsafe.
Printing out an infinite list of Doubles with mapM_ print is perfectly fine; the list will be processed incrementally and the program should run with constant memory usage. I suspect there is a bug in the System.Random.Mersenne module you're using, or a bug the C library it's based on, or a problem with your computer (such as faulty RAM).1 Note that newMTGen comes with this warning:
Due to the current SFMT library being vastly impure, currently only a single generator is allowed per-program. Attempts to reinitialise it will fail.
You might be better off using the provided global MTGen instead.
That said, you can't convert IO [Double] into [IO Double] in that way; there's no way to know how long the resulting list would be without executing the IO action, which is impossible, since you have a pure result (albeit one that happens to contain IO actions). For infinite lists, you could write:
desequence :: IO [a] -> [IO a]
desequence = desequence' 0
where
desequence n m = fmap (!! n) m : desequence (n+1) m
But every time you execute an action in this list, the IO [a] action would be executed again; it'd just discard the rest of the list.
The reason randoms can work and return an infinite list of random numbers is because it uses lazy IO with unsafeInterleaveIO. (Note that, despite the "unsafe" in the name, this one can't cause segfaults, so something else is afoot.)
1 Other, less likely possibilities include a miscompilation of the C library, or a bug in GHC.
Suppose I wanted to print out only the first ten values of xs. How could I do that?
Just use take:
main =
do g <- (newMTGen Nothing)
xs <- (randoms g) :: IO [Double]
mapM_ print $ take 10 xs
You wrote
xs has type IO [Double]
But actually, randoms g has type IO [Double], but thanks to the do notation, xs has type [Double], you can just apply take 10 to it.
You could also skip the binding using liftM:
main =
do g <- newMTGen Nothing
ys <- liftM (take 10) $ randoms g :: IO [Double]
mapM_ print ys

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