Types and do notation

4
votes

This is several questions rolled into one:

  1. In do notation, does each line have to return the same type? For example, can I write one line in a single do block that returns an IO monad, and another that returns an integer? (My understanding, based on how the de-sugaring with >> and >>= seems to work, is that the answer is no.)

  2. If not, then how does the compiler determine what type the lines must all return? In all the examples I've seen, the author takes it as a foregone conclusion that we're just working with IO monads. But how do you know, for a given do block, what each line must return?

  3. Again assuming the answer to #1 is no: How do you use functions that don't return the right kind of monad inside a do block? For example, consider this websockets code:

    application :: MVar ServerState -> WS.Request -> WS.WebSockets WS.Hybi00 ()
application state rq = do
  WS.acceptRequest rq
  msg <- WS.receiveData :: WS.WebSockets WS.Hybi00 Text
  return ()

    Suppose I want to print the value of msg. How would I go about that in a way that doesn't conflict with the type of the do block?

2
haskellmonads
Why don't you just try it? Further, since do-notation is just syntactic sugar (which, as you said. translates into (>>=) and (>>)) the mechanism at hand is exactly the same as for any other type class and not specific to do-notation and Monads. - chris
Sorry, I accidentally messed up the code formatting with my last edit (that introduced the enumeration). Unfortunately, my correction would have too few characters to be allowed as an edit. Could you remedy this again, please? - chris
Yes, I fixed it. I see how you tried to use the Markdown numbered lists, but they appear to conflict with the code block. Which is why I chose to put each number in its own paragraph. - rlkw1024
You would have to indent the code block a further 4 spaces (i.e., 8 in total) to make it work inside the enumeration, I think. - chris

2 Answers

9
votes
  1. In a do-block, each line can return different types, but they must be in the same monad
    • One line can return an IO String and one can return an IO Integer but they both have to be IO.
  2. The same as the rest of Haskell. Type inference. Just like in the rest of Haskell it doesn't always work, and when it doesn't you must annotate as well.
  3. There are 2 ways of doing this
    • let, remember how in GHCi you must use let to declare local variables, you can do the same in a do block. let someMonad = doSomething
      • Notice, no in
    • Monad Transformers! This to big a topic to explain in a blurb, but basically they're monads that have a special function lift which can "lift" another monad into the transformer. Transformers normally end in a T, eg StateT. Pretty much every monad you use has an equivalent transformer.
4
votes

To answer the final part of your final question,

Suppose I want to print the value of msg. How would I go about that in a way that doesn't conflict with the type of the do block?

As jozefg said in his answer, monad transformers is usually what you need for this. However, in this case the WebSockets p monad is not a transformer. But, it is an instance of MonadIO which is a class for monad stacks which have IO "at the bottom" and which therefore let you run arbitrary IO actions from within them.

The MonadIO class provides the function liftIO which has the type

liftIO :: MonadIO m => IO a -> m a

In your case, this becomes IO a -> WebSockets Hybi00 a, so you can use it to convert the print msg action from IO () to WebSockets Hybi00 (), which you can then use in the do block:

application :: MVar ServerState -> WS.Request -> WS.WebSockets WS.Hybi00 ()
application state rq = do
  WS.acceptRequest rq
  msg <- WS.receiveData :: WS.WebSockets WS.Hybi00 Text
  liftIO $ print msg