Hi,
I am using the System.Random method randomRIO. How can I convert its output to an Int? Thanks... 
You can not convert an IO Int to Int, or at least, you shouldn't.
However, you can do as follows: test :: IO () test = do int < randomRIO  or whatever it is called print $ useInt int useInt :: Int > Int useInt x = x+10 //Tobias 2009/6/9 ptrash <[hidden email]>: > > Hi, > > I am using the System.Random method randomRIO. How can I convert its output > to an Int? > > Thanks... >  > View this message in context: http://www.nabble.com/ConvertIOInttoInttp23940249p23940249.html > Sent from the Haskell  HaskellCafe mailing list archive at Nabble.com. > > _______________________________________________ > HaskellCafe mailing list > [hidden email] > http://www.haskell.org/mailman/listinfo/haskellcafe >  Tobias Olausson [hidden email] _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by ptrash
ptrash wrote:
> Hi, > > I am using the System.Random method randomRIO. How can I convert its output > to an Int? > > Thanks... You cannot [1], you should read up on monads and I/O in Haskell, for example http://haskell.org/haskellwiki/IO_inside [1] Yes, you can, but no, you don't want to. Regards,  Jochem Berndsen  [hidden email] GPG: 0xE6FABFAB _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by Tobias Olausson
On 2009/06/09, at 19:33, Tobias Olausson wrote: > You can not convert an IO Int to Int, or at least, you shouldn't. > However, you can do as follows: > > test :: IO () > test = do > int < randomRIO  or whatever it is called > print $ useInt int > > useInt :: Int > Int > useInt x = x+10 Or, you can lift pure function into IO. the below test' function almost same as above test function. (But I used randomIO instead of randomRIO because it seemed to be a typo :) test' = print =<< fmap useInt randomIO I think it is more handy than using do notation, when you want to do something simple with monads. And converting IO Int to IO anything is much easier and safer than converting IO Int to Int. ghci> :m +System.Random Data.Char ghci> :t fmap (+1) randomIO fmap (+1) randomIO :: (Num a, Random a) => IO a ghci> :t fmap show randomIO fmap show randomIO :: IO String ghci> :t fmap chr randomIO fmap Data.Char.chr randomIO :: IO Char ghci> :t fmap (+) randomIO fmap (+) randomIO :: (Num a, Random a) => IO (a > a) Thanks, Hashimoto > //Tobias > > 2009/6/9 ptrash <[hidden email]>: >> >> Hi, >> >> I am using the System.Random method randomRIO. How can I convert >> its output >> to an Int? >> >> Thanks... >>  >> View this message in context: http://www.nabble.com/ConvertIOInt >> toInttp23940249p23940249.html >> Sent from the Haskell  HaskellCafe mailing list archive at >> Nabble.com. >> >> _______________________________________________ >> HaskellCafe mailing list >> [hidden email] >> http://www.haskell.org/mailman/listinfo/haskellcafe >> > > > >  > Tobias Olausson > [hidden email] > _______________________________________________ > HaskellCafe mailing list > [hidden email] > http://www.haskell.org/mailman/listinfo/haskellcafe _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by ptrash
Ok, thanks for the information.

In reply to this post by ptrash
Hmm...it am not getting through it. I just want to generate a random number and then compare it with other numbers. Something like
r = randomRIO (1, 10) if (r > 5) then... else ... 
On Tue, Jun 9, 2009 at 2:52 PM, ptrash<[hidden email]> wrote:
> > Hmm...it am not getting through it. I just want to generate a random number > and then compare it with other numbers. Something like > > r = randomRIO (1, 10) > if (r > 5) then... else ... You have to do it inside the IO monad, something like myFunc = do r < randomRIO (1, 10 if r > 5 then ... else ... /M  Magnus Therning (OpenPGP: 0xAB4DFBA4) magnus＠therning．org Jabber: magnus＠therning．org http://therning.org/magnus identi.catwitter: magthe _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by ptrash
On Tue, 9 Jun 2009, ptrash wrote: > I am using the System.Random method randomRIO. How can I convert its output > to an Int? in general: http://haskell.org/haskellwiki/How_to_get_rid_of_IO about randomIO: http://haskell.org/haskellwiki/Avoiding_IO#State_monad _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by Magnus Therning
Am Dienstag 09 Juni 2009 15:57:24 schrieb Magnus Therning:
> On Tue, Jun 9, 2009 at 2:52 PM, ptrash<[hidden email]> wrote: > > Hmm...it am not getting through it. I just want to generate a random > > number and then compare it with other numbers. Something like > > > > r = randomRIO (1, 10) > > if (r > 5) then... else ... > > You have to do it inside the IO monad, something like > > myFunc = do > r < randomRIO (1, 10 > if r > 5 > then ... > else ... > > /M Or make the source of the pseudorandom numbers explicit: import System.Random function :: (RandomGen g, Random a) => g > other args > result function gen whatever  r > 5 = blah newgen something  r < 3 = blub newgen somethingElse  otherwise = bling where (r,newgen) = randomR (lo,hi) gen and finally, when the programme is run: main = do args < getArgs sg < getStdGen foo < thisNThat print $ function sg foo If you're doing much with random generators, wrap it in a State monad. _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
On Tue, Jun 9, 2009 at 16:14, Daniel Fischer<[hidden email]> wrote:
> Am Dienstag 09 Juni 2009 15:57:24 schrieb Magnus Therning: >> On Tue, Jun 9, 2009 at 2:52 PM, ptrash<[hidden email]> wrote: >> > Hmm...it am not getting through it. I just want to generate a random >> > number and then compare it with other numbers. Something like >> > >> > r = randomRIO (1, 10) >> > if (r > 5) then... else ... >> >> You have to do it inside the IO monad, something like >> >> myFunc = do >> r < randomRIO (1, 10 >> if r > 5 >> then ... >> else ... >> >> /M > > Or make the source of the pseudorandom numbers explicit: > > import System.Random > > function :: (RandomGen g, Random a) => g > other args > result > function gen whatever >  r > 5 = blah newgen something >  r < 3 = blub newgen somethingElse >  otherwise = bling > where > (r,newgen) = randomR (lo,hi) gen > > and finally, when the programme is run: > > main = do > args < getArgs > sg < getStdGen > foo < thisNThat > print $ function sg foo > > If you're doing much with random generators, wrap it in a State monad. To avoid reinventing the wheel one can use excellent package available on Hackage: http://hackage.haskell.org/cgibin/hackagescripts/package/MonadRandom > The die function simulates the roll of a die, picking a number between 1 and 6, inclusive, and returning it in the Rand monad. > Notice that this code will work with any source of random numbers g. > > die :: (RandomGen g) => Rand g Int > die = getRandomR (1,6) > > The dice function uses replicate and sequence to simulate the roll of n dice. > > dice :: (RandomGen g) => Int > Rand g [Int] > dice n = sequence (replicate n die) > > To extract a value from the Rand monad, we can can use evalRandIO. > > main = do > values < evalRandIO (dice 2) > putStrLn (show values) Best regards Krzysztof Skrzętnicki _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by Magnus Therning
Magnus Therning writes:
> ptrash wrote: >> ...am not getting through it. I just want to generate a random number >> and then compare it with other numbers. Something like >> >> r = randomRIO (1, 10) >> if (r > 5) then... else ... > > You have to do it inside the IO monad, something like > > myFunc = do > r < randomRIO (1, 10 This may continue forever... With nice references to monads, to Unsafe@#*!, etc. ... We may say, as many tutorials do : "this is not what you want!" (which I hate ; you are not my conscience, whoever you are...), or just give some code, not always readable... Perhaps I belong to a minority here, but I strongly believe that at THIS level, the first thing to do  unless I am dead wrong  is to explain to our friend ptrash (who could find a less gothic pseudo) that in a pure functional programming, the construction r = whatEver(par1,par2) being a function call, cannot give "just a random number", something which is not (intuitively) determined, and changes with every call, despite the constancy of the arguments. For most of us, acquainted with the stuff, it becomes trivial, but if somebody doesn't know that a classical pseudorandom generator modifies a "seed", and in such a way involves a "side effect", then sending him to the monadic heaven is dangerous. Please, tell him first about random streams, which he can handle without IO. Or, about ergodic functions (hashing contraptions which transform ANY parameter into something unrecognizable). When he says : "I know all that", THEN hurt him badly with monads. Jerzy Karczmarczuk _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
Hello jerzy,
Tuesday, June 9, 2009, 8:23:04 PM, you wrote: > Please, tell him first about random streams, which he can handle without > IO. Or, about ergodic functions (hashing contraptions which transform ANY > parameter into something unrecognizable). When he says : "I know all that", > THEN hurt him badly with monads. i think that for someone coming from imperative programming teeling about IO monad is the easiest way. and then he will learn how to do it FP way  Best regards, Bulat mailto:[hidden email] _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
Bulat Ziganshin wrote:
> Hello jerzy, > > Tuesday, June 9, 2009, 8:23:04 PM, you wrote: > > >> Please, tell him first about random streams, which he can handle without >> IO. Or, about ergodic functions (hashing contraptions which transform ANY >> parameter into something unrecognizable). When he says : "I know all that", >> THEN hurt him badly with monads. >> > > i think that for someone coming from imperative programming teeling > about IO monad is the easiest way. and then he will learn how to do it > FP way > help me at all back then. At least in the beginning you want to detach yourself from an imperative style, not try to simulate it with some weird structure that you don't really understand. More generally I really wish IO hadn't been the first Monad I played with. It's so close to a Functor, yet in my mind Functors were simple, just structures that could be mapped, and Monads were these mysterious things that allowed you to get away with side effects and that once you were inside you could never get out. Jorge _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by Krzysztof Skrzętnicki
2009/6/9 Krzysztof Skrzętnicki <[hidden email]>
Please do! Prefer MonadRandom to explicit generator passing: http://lukepalmer.wordpress.com/2009/01/17/usemonadrandom/. Keep computations in MonadRandom, and pull them out with evalRandomIO at the last second.
Luke
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In reply to this post by Bulat Ziganshin2
On Tue, 9 Jun 2009, Bulat Ziganshin wrote: > Hello jerzy, > > Tuesday, June 9, 2009, 8:23:04 PM, you wrote: > >> Please, tell him first about random streams, which he can handle without >> IO. Or, about ergodic functions (hashing contraptions which transform ANY >> parameter into something unrecognizable). When he says : "I know all that", >> THEN hurt him badly with monads. > > i think that for someone coming from imperative programming teeling > about IO monad is the easiest way. and then he will learn how to do it > FP way I came from imperative programming and never wanted to use randomIO, because it forces you to IO and randomsIO is not lazy. _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by ptrash
Hi,
I have tried on the console to write x < randomRIO(1,10) :t x Everythings fine and the type of x is x :: Integer Now I have tried to write a Method which gives me a Number of random numbers the same way but it doesn't work. randomList :: Int > [Integer] randomList 0 = [] randomList n = do r < randomRIO (1, 10) r:randomList(n1) It says Couldn't match expected type `IO t' against inferred type `[t]' r < randomRIO (1,10) causes an error. But why does it work on the console? Is there a way to solve it another way? 
"r < randomRIO (1,10)" is NOT the source of error. Why do you think it is?
ptrash wrote on 10.06.2009 15:55: > Hi, > > I have tried on the console to write > > x < randomRIO(1,10) > :t x > > Everythings fine and the type of x is > x :: Integer > > Now I have tried to write a Method which gives me a Number of random numbers > the same way but it doesn't work. > > randomList :: Int > [Integer] > randomList 0 = [] > randomList n = do > r < randomRIO (1, 10) > r:randomList(n1) > > It says Couldn't match expected type `IO t' against inferred type `[t]' > r < randomRIO (1,10) causes an error. But why does it work on the console? > Is there a way to solve it another way? HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by ptrash
Hmm...I use the Eclipse Plugin. And this row is marked as error. Then where is the error?

In reply to this post by ptrash
On 10 Jun 2009, at 12:55 pm, ptrash wrote:
> > > Now I have tried to write a Method which gives me a Number of > random numbers > the same way but it doesn't work. > > randomList :: Int > [Integer] > randomList 0 = [] > randomList n = do > r < randomRIO (1, 10) > r:randomList(n1) > > It says Couldn't match expected type `IO t' against inferred type > `[t]' > r < randomRIO (1,10) causes an error. But why does it work on the > console? > Is there a way to solve it another way? > I had the same problem a while back, the thread is here http://www.mailarchive.com/haskellcafe@.../msg46194.html the console uses IO already, so it's not a problem there. I ended up learning about the >>= operator, and that helped a lot. Anyway, lots of helpful links in that mail thread. Iain _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
In reply to this post by ptrash
On Wed, Jun 10, 2009 at 12:55 PM, ptrash <[hidden email]> wrote:
The type of x *in the context of an IO computation* is Integer. GHCi is basically an IO computation.
Another example:
foo :: Integer > Integer
foo x = x+1
main :: IO ()
main = do
x < randomRIO (1,10)
print (foo x)
This is fine. In the context of the IO computation "main", x is bound to the result of "randomRIO (1,10)", and you can pass it to functions expecting Integer values (not IO Integer!). So in this way, and this way only, you can access the Integer returned by an IO action. You can *not* access the Integer returned by an IO action from within a normal function, *only* by by binding it to a variable (with "<") inside *another IO action*.
I'm not sure what text you're using to learn Haskell, but a very basic and fundamental property of Haskell (and indeed 99% of why it's cool, IMO) is that code which does side effects (like reading from a global random number seed), and code which does not do side effects (i.e. functions which always return the same result given the same input) are kept extremely separate. This appears to be the source of your confusion. It's simply not possible to do side effect inside a normal function, just like it's not possible to cast an arbitrary integer to a pointer in Java  the language is designed to not require it, and the benefits of being able to trust that your program obeys certain properties are worth it.
>randomList :: Int > [Integer]
>randomList 0 = [] >randomList n = do > r < randomRIO (1, 10) > r:randomList(n1) In this code you're trying to do side effects from within a pure function. This is *not* allowed. You must either make randomList an IO action (i.e returning IO [Integer]), or remove any impurities from its implementation. For example you can make randomList take a randon number generator and use the random number generator to produce random values: randomList :: (RandomGen g) > Int > g > [Integer]
randomList 0 _ = []
randomList n generator = r : randomList (n1) newGenerator
where (r, newGenerator) = randomR (1,10) generator
This is totally pure, since if you pass in the same random number generator multiple times, you'll get the exact same
result. Note that randomR returns a random values and a new random number generator (you wouldn't want to pass along the same one in the recursive call to randomList as that would give you an identical random number each time you use it!).
So where do you get the random number generator from? Well one way is to make your own using "mkStdGen", which produces one from a seed (it will give you an identical one given an identical seed). Another way is to use "newStdGen" to generate one from within an IO action:
main :: IO ()
main = do
generator < newStdGen
print ( randomList 10 generator )
The point, though, is that things having side effects (such as newStdGen) can only be used in the context of something else having side effects. So the "IO" type is "contagious", as soon as you use it in a function, then that function must also return IO, and so on for anything using *that* function and son.
 Sebastian Sylvan +44(0)7857300802 UIN: 44640862 _______________________________________________ HaskellCafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskellcafe 
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