输入IO的haskell说明

Question

Can any one help me with understanding this function type? stateOfMind :: BotBrain -> IO (Phrase -> Phrase)

stateOfMind is a function and BotBrain is just a type.

type Phrase = [String]
type PhrasePair = (Phrase, Phrase)
type BotBrain = [(Phrase, [Phrase])]

If stateOfMind had this type: BotBrain -> (Phrase -> Phrase) , then stateOfMind would take a BotBrain as an argument to produce a new function that takes a Phrase and gives a Phrase as a result. But now we have an IO , ie IO (Phrase -> Phrase) . What does that mean?

randomIO has monadic type but why is to like that? Is is because of the seed we choose? Monadic I normally used for input and output, but a random generator actually doesn't get any input from user at run time.

Answer 1

stateOfMind :: BotBrain -> IO (Phrase -> Phrase)

stateOfMind takes a BotBrain and returns an IO action whose result is a function from Phrase to Phrase . When you execute an IO (Phrase -> Phrase) action in the IO monad, you get an ordinary pure function Phrase -> Phrase .

randomIO is in IO because it uses getStdRandom to change the state of the global standard random number generator. Multiple calls to randomIO can (and will) return different results, which is precisely what IO indicates.

You can create a pure generator with mkStdGen and get values from it purely:

let
  g1 = mkStdGen 42      -- Seed value
  (r1, g2) = random g1  -- Get first random number
  (r2, g3) = random g2  -- Get second random number
in r1 + r2

And a convenient way to get rid of the repetition here is to use the State monad:

flip evalState (mkStdGen 42) $ do
  r1 <- state random
  r2 <- state random
  return (r1 + r2)