On Mon, Jul 30, 2007 at 02:40:35PM -0700, Chad Scherrer wrote:
I'm trying to do something I thought would be pretty simple, but it's giving me trouble.
Given a list, say [1,2,3], I'd like to be able to generate an infinite list of random elements from that list, in this case maybe [1,2,1,3,2,1,3,2,3,1,2,...]. I'm using IO for random purely due to laziness (my own, not Haskell's).
I was thinking the best way to do this might be to first write this function:
randomElts :: [a] -> [IO a] randomElts [] = [] randomElts [x] = repeat (return x) randomElts xs = repeat r where bds = (1, length xs) xArr = listArray bds xs r = do i <- randomRIO bds return (xArr ! i)
Then I should be able to do this in ghci:
sequence . take 5 $ randomElts [1,2,3] [*** Exception: stack overflow
Any idea what's going on? I thought laziness (Haskell's, not my own) would save me on this one.
The code you posted works fine for me (GHCi 6.7.20070712). However, it's pretty bad style in a way that suggests a misunderstanding of IO. A value of type IO t is not a "tainted t", it's an "action that returns t". So, in general: do let xv = randomRIO (1,20) a <- xv b <- xv return (a == b) will normally return *False*. Why? Because, while the same action is used, it doesn't always return the same value! In general, when you see a type of the form [IO a] or Maybe (IO a) or IO a -> b, ask yourself if that's what you really want. (Sometimes it is, and the flexibility of having IO anywhere is very powerful). A better way to write randomElts is: randomElt :: [a] -> IO a randomElt xs = do ix <- randomRIO bds return (arr ! i) where arr = listArray bds xs bds = (1, length xs) randomElts :: Int -> [a] -> IO [a] randomElts n xs = replicateM n (randomElt xs) -- uses replicateM from Control.Monad Sadly, it's very hard to "just return an infinite list" in IO. it can be done in dedicated monads, however. Stefan