Hello
I am studying Real World Haskell chapter 9. Here is a snippet of code

data Info = Info {
      infoPath :: FilePath
    , infoPerms :: Maybe Permissions

    , infoSize :: Maybe Integer
    , infoModTime :: Maybe ClockTime
    } deriving (Eq, Ord, Show)

getInfo :: FilePath -> IO Info

-- file: ch09/ControlledVisit.hs

traverse order path = do
    names <- getUsefulContents path
    contents <- mapM getInfo (path : map (path </>) names)
    liftM concat $ forM (order contents) $ \info -> do
      if isDirectory info && infoPath info /= path

        then traverse order (infoPath info)
        else return [info]

getUsefulContents :: FilePath -> IO [String]
getUsefulContents path = do
    names <- getDirectoryContents path
    return (filter (`notElem` [".", ".."]) names)


isDirectory :: Info -> Bool
isDirectory = maybe False searchable . infoPerms

When I read about IO in the previous chapter, I learnt that reading a file can be done lazily.
Here my doubt is that whether the expression "order contents" would generate a list of directory contents (held in memory) for use in forM construct. Because in a subsequent para, I find this line.

"If we are traversing a directory containing 100,000 files of which we care about three, we'll allocate a 100,000-element list before we have a chance to trim it down to the three we really want"


Wouldn't laziness ensure that in the traverse function, when iterating over directory contents using the list generated by "order contents",
it will generate just one element at a time and then free the memory for that entry immediately since we are not using the referencing list 

anymore in the rest of the function. 

Not sure whether I have understood the concept of laziness w.r.t monads correctly. Please clarify my doubts with regard to the code snippet.

Thanks for your time.

--
Regards
Lakshmi Narasimhan T V