Haskell: Something like the application `$` operator for "do" notation?

Question

I'm supplying a function to idleCallback This notation works:

idleCallback $= Just (do
    modifyIORef world play
    postRedisplay Nothing)

Why doesn't this (seemingly similar) notation work?

idleCallback $= Just $ do
    modifyIORef world play
    postRedisplay Nothing

To save your hoogling, the types are:

($=) :: HasSetter s => s a -> a -> IO ()
type IdleCallback = IO ()
data SettableStateVar a 
idleCallback :: SettableStateVar (Maybe IdleCallback)
postRedisplay :: Maybe Window -> IO ()
modifyIORef :: IORef a -> (a -> a) -> IO ()

GHC says:

Couldn't match expected type `Maybe IdleCallback'
            with actual type `a0 -> Maybe a0'
In the second argument of `($=)', namely `Just'
In the expression: idleCallback $= Just
In a stmt of a 'do' block:
  idleCallback $= Just
  $ do { modifyIORef world play;
         postRedisplay Nothing }

Can this be written without wrapping the do block in parenthesis?

Answer 1

It is a precedence error.... ($=) is binding more tightly than ($). You can see this in the error message:

Couldn't match expected type `Maybe IdleCallback'
        with actual type `a0 -> Maybe a0'
In the second argument of `($=)', namely `Just'

It thinks that the second argument of ($=) is simply Just (which is a valid Haskell type referencing a function). If you put parenthesis around the whole Just, including the do-block, it should work.