Haskell help, /prog/!


-- create a random tree.
randomTree :: Random r => IO (BinaryTree r)
randomTree =
     -- randomly pick the left and right sides.
  do leftIO  <- pick(randomLeaf, randomBranch)
     rightIO <- pick(randomLeaf, randomBranch)
     -- "un-IO" the sides.
     left    <- leftIO
     right   <- rightIO
     -- get a random value for the top of the branch/tree.
     value   <- randomIO
     -- create branch and return.
     return (Branch left value right)
  where
    -- return left or right at random.
    pick (left, right) =
      do useLeft <- randomRIO(True, False)
         if useLeft then return left else return right
    -- create a random leaf.
    randomLeaf =
      do value <- randomIO
         let leaf = Leaf value
         return leaf
    -- setup an alias.
    randomBranch = randomTree

This code actually works fine, but I'm wondering if there's a way to simplify the "leftIO/rightIO" stuff. Is there some sort of "double arrow" that will un-Monad twice, or is this verbosity unavoidable?

>>20
In randomBranch you have 0.5 chance of terminating and 0.5 chanche of calling randomIO recursively. randomIO calls randomBranch two times so on average one branch will terminate and the other continue. So it will run forever.