Risks of using unsafeperformIO on randomIO

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I am creating a Haskell application that generates a random number on an infinite loop (only when requested by a client). However, I should only use pure functions for that purpose. Is it safe to wrap randomIO with unsafeperformIO without any drastic stability or performance risk?

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Joachim Breitner On BEST ANSWER

Any use of unsafePerformIO should be justified by a proof that the resulting value is still pure. The rigour of the proof is up to you and the importance of the work. For example, this pathetic use unsafePerformIO and randomIO should be safe, because you can prove that when slowTrue returns anything, it will return True.

import System.Random
import System.IO.Unsafe
import Data.Int

slowTrue = unsafePerformIO $ go
  where
    go = do
        x1 <- randomIO
        x2 <- randomIO
        if ((x1 :: Int16) == x2) then return True else go

The following tempting definition of a global, possibly random variables is not safe:

rand :: Bool -> Int
rand True = unsafePerformIO randomIO 
rand False = 0

The problem is that the same expression will now yield different values:

main = do
    print (rand True)
    print (rand True)

prints here:

-7203223557365007318
-7726744474749938542

(at least when compiled without optimizations – but that just stresses the fragility of inappropriate use of unsafePerformIO).

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I am creating a Haskell application that generates a random number on an infinite loop (only when requested by a client). However, I should only use pure functions for that purpose. Is it safe to wrap randomIO with unsafeperformIO without any drastic stability or performance risk?

Aside from performance (no risk) or stability (small risk), consider also the fact that you are writing non-canonical Haskell for no good reason. Your code will be difficult for anyone else to maintain if you take this approach.