Haskell - Calculation of the time in the monad Rand

Question

I want to evaluate random calculations in Haskell with a timeout using the Control.Monad.Random library. The following works fine:

 ghci> import System.Timeout ghci> import Control.Monad.Random ghci> timeout 1000 . evalRandIO $ getRandomR (True, False) Just True 

However, this approach does not work if we have a calculation of the type Rand StdGen a that never ends (i.e. is evaluated from below). For instance:

 ghci> let f = f :: Rand StdGen Bool ghci> timeout 1000 $ evalRandIO f Just 

Here GHCI prints "Just" and then hangs endlessly to evaluate f . Can someone who knows more about Haskell runtime than I explain why this happens and how to get around it? I assume that the evalRandIO f expression evaluates to WHNF, so timeout 10 thinks the calculation will end, but I really don't know.

Answer 1

It might make sense if you have to do something like

 > Just x <- timeout 1000 $ evalRandIO f > :tx x :: Bool > x Interrupted. 

The calculation itself is completed, namely, reaches WHNF, so timeout does not catch it. The timeout 1000 function itself ends and returns Just undefined . An example where you can get a timeout to catch a lower bound would be

 > import Control.DeepSeq > :set +m -- Multiline mode > let f :: Bool | f = f | > timeout 1000000 $ deepseq f (return f) Nothing 

You will see that it freezes for a second, and then returns Nothing when deepseq does not finish evaluating f , so it can execute return f .

So your problem is that f = f gets a WHNF score instead of NF. To force NF, you need to use something like deepseq . Another, perhaps simpler example would be to simply use the $! operator $! , eg:

 > let f :: Rand StdGen Bool | f = f | > timeout 1000000 $ evalRandIO $! f Nothing