Re: [Haskell-cafe] Convert IO Int to Int
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(n-1) 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? -- View this message in context: http://www.nabble.com/Convert-IO-Int-to-Int-tp23940249p23960652.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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(n-1) 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? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
Hmm...I use the Eclipse Plugin. And this row is marked as error. Then where is the error? -- View this message in context: http://www.nabble.com/Convert-IO-Int-to-Int-tp23940249p23960827.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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(n-1) 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.mail-archive.com/haskell-cafe@haskell.org/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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
On Wed, Jun 10, 2009 at 12:55 PM, ptrash ptr...@web.de wrote: 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 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(n-1) 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 (n-1) 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)7857-300802 UIN: 44640862 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
On Wed, Jun 10, 2009 at 2:08 PM, Sebastian Sylvan sebastian.syl...@gmail.com wrote: randomList :: (RandomGen g) - Int - g - [Integer] Just spotted this typo, it should be: randomList :: (RandomGen g) = Int - g - [Integer] There may be other minor typos as I don't have a compiler handy. -- Sebastian Sylvan +44(0)7857-300802 UIN: 44640862 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
On Wed, Jun 10, 2009 at 2:10 PM, Sebastian Sylvan sebastian.syl...@gmail.com wrote: On Wed, Jun 10, 2009 at 2:08 PM, Sebastian Sylvan sebastian.syl...@gmail.com wrote: randomList :: (RandomGen g) - Int - g - [Integer] Just spotted this typo, it should be: randomList :: (RandomGen g) = Int - g - [Integer] There may be other minor typos as I don't have a compiler handy. Oh come on! randomList :: (RandomGen g) = Int - g - [Integer] -- Sebastian Sylvan +44(0)7857-300802 UIN: 44640862 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
This stuff is tricky for most newcomers I suspect (it was for me) x - randomRIO(1,10) is temporarilly pulling the Integer you've named x out of the IO Integer it came from. You can think of the console as being an input/output stream inside the IO monad, which is why it is allowed there. The fact is these are equivalent do x - randomRIO(1,10) x : expression and randomRIO(1,10) = (\x - x:expression) They are the same meaning, and both are illegal for the same reason. (=) is the bind operator for Monads and it allows you to do things like sequence an IO operation or action, then examine the contents returned by that IO action, possibly performing some transformation on it with another function. x:expression is the function that takes a value x and applies the cons function (:) to build a list with expression. The problem is that the type system of Haskell will not allow this because the bind function (which is used by the do syntax behind the scenes) is of type: (=) :: (Monad m) = m a - (a - m b) - m b Which states that the first argument to (=) is a Monad a, or in your case randomRIO(1,10) which is of type IO Integer IO being the m part of m a and Integer being the a part of m a. What comes next is your consing function of (x : expression). This is of type [Integer]. So you failed to give it a function of type (a - m b). (a - m b) says that the input type to that function must be the same as the contained value of the first argument to (=) (because they're both of type 'a'). The m b part says that you must use the *same Monad* you used in the first parameter to (=) which is IO, not []. However the 'b' part says you can convert things of one type 'a' to things of another type 'b'. This is legal: randomRIO(1,10) = return (x : expression) However what you've got now is not a List of Integers ([Integer]) but an IO [Integer]. Because of the type of (=), you are not going to ever permanently escape that IO wrapper around your values you're interested in. The return is a function of the class Monad, that takes a value, and wraps it up in a monad. Based on the (=) function's type, the system can infer that you meant the IO monad here. You can even test this at the console: Prelude System.Random randomRIO(1,10) = (\x - return (x:[99])) [7,99] Prelude System.Random randomRIO(1,10) = (\x - return (x:[99])) [6,99] Prelude System.Random randomRIO(1,10) = (\x - return (x:[99])) [2,99] Prelude System.Random :t randomRIO(1,10) = (\x - return (x:[99])) randomRIO(1,10) = (\x - return (x:[99])) :: (Num t, Random t) = IO [t] vs randomRIO(1,10) = (\x - x:[99]) Which doesn't pass the type checking of the system because [] is not the same monad as IO. Perhaps you'd do well for yourself by reading Learn You A Haskell? I've found it to be pretty darned good at explaining lots of haskell to newcomers. Dave On Wed, Jun 10, 2009 at 4:55 AM, ptrash ptr...@web.de wrote: 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(n-1) 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? -- View this message in context: http://www.nabble.com/Convert-IO-Int-to-Int-tp23940249p23960652.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
Thanks a lot. I have put now everything into the main method and it works. -- View this message in context: http://www.nabble.com/Convert-IO-Int-to-Int-tp23940249p23964365.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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 ptr...@web.de: 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/Convert-IO-Int-to-Int-tp23940249p23940249.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- Tobias Olausson tob...@gmail.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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 | joc...@functor.nl GPG: 0xE6FABFAB ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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 ptr...@web.de: 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/Convert-IO-Int- to-Int-tp23940249p23940249.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- Tobias Olausson tob...@gmail.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
Ok, thanks for the information. -- View this message in context: http://www.nabble.com/Convert-IO-Int-to-Int-tp23940249p23942344.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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 ... -- View this message in context: http://www.nabble.com/Convert-IO-Int-to-Int-tp23940249p23943301.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
On Tue, Jun 9, 2009 at 2:52 PM, ptrashptr...@web.de 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.ca|twitter: magthe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
Am Dienstag 09 Juni 2009 15:57:24 schrieb Magnus Therning: On Tue, Jun 9, 2009 at 2:52 PM, ptrashptr...@web.de 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 pseudo-random 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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
On Tue, Jun 9, 2009 at 16:14, Daniel Fischerdaniel.is.fisc...@web.de wrote: Am Dienstag 09 Juni 2009 15:57:24 schrieb Magnus Therning: On Tue, Jun 9, 2009 at 2:52 PM, ptrashptr...@web.de 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 pseudo-random 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/cgi-bin/hackage-scripts/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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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 uns...@#*!, 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 pseudo-random 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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
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 a imperative programming background. I didn't feel like this 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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Convert IO Int to Int
2009/6/9 Krzysztof Skrzętnicki gte...@gmail.com On Tue, Jun 9, 2009 at 16:14, Daniel Fischerdaniel.is.fisc...@web.de wrote: 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/cgi-bin/hackage-scripts/package/MonadRandom Please do! Prefer MonadRandom to explicit generator passing: http://lukepalmer.wordpress.com/2009/01/17/use-monadrandom/. Keep computations in MonadRandom, and pull them out with evalRandomIO at the last second. Luke http://hackage.haskell.org/cgi-bin/hackage-scripts/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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
