You can wrap values into actions and pipe the result of one computation into another:

return :: Monad m => a -> m a
(>>=)  :: Monad m => m a -> (a -> m b) -> m b

However, the definition of a Monad doesn’t guarantee the existence of a function of type Monad m => m a -> a.

That means there is, in general, no way to extract a value from a computation (i.e. “unwrap” it). This is the case for many instances:

extract :: Maybe a -> a
extract (Just x) = x          -- Sure, this works, but...
extract Nothing  = undefined  -- We can’t extract a value from failure.

Specifically, there is no function IO a -> a, which often confuses beginners; see this example (opens new window).

As of GHC 7.10, Applicative is a superclass of Monad (i.e., every type which is a Monad must also be an Applicative). All the methods of Applicative (pure, <*>) can be implemented in terms of methods of Monad (return, >>=).

It is obvious that pure and return serve equivalent purposes, so pure = return. The definition for <*> is too relatively clear:

mf <*> mx = do { f <- mf; x <- mx; return (f x) }                 
       -- = mf >>= (\f -> mx >>= (\x -> return (f x)))
       -- = [r   | f <- mf, x <- mx, r <- return (f x)]   -- with MonadComprehensions
       -- = [f x | f <- mf, x <- mx]                   

This function is defined as ap in the standard libraries.

Thus if you have already defined an instance of Monad for a type, you effectively can get an instance of Applicative for it "for free" by defining

instance Applicative < type > where
    pure  = return
    (<*>) = ap

As with the monad laws, these equivalencies are not enforced, but developers should ensure that they are always upheld.

class Monad m where
    return :: a -> m a
    (>>=) :: m a -> (a -> m b) -> m b

The most important function for dealing with monads is the bind operator >>=:

(>>=) :: m a -> (a -> m b) -> m b

• Think of m a as "an action with an a result".
• Think of a -> m b as “an action (depending on an a parameter) with a b result.”.

>>= sequences two actions together by piping the result from the first action to the second.

The other function defined by Monad is:

return :: a -> m a

Its name is unfortunate: this return has nothing to do with the return keyword found in imperative programming languages.

return x is the trivial action yielding x as its result. (It is trivial in the following sense (opens new window)????

return x >>= f       ≡  f x     --  “left identity” monad law
       x >>= return  ≡  x       -- “right identity” monad law

Maybe is used to represent possibly empty values - similar to null in other languages. Usually it is used as the output type of functions that can fail in some way.

Consider the following function:

halve :: Int -> Maybe Int
halve x
  | even x = Just (x `div` 2)
  | odd x  = Nothing

Think of halve as an action, depending on an Int, that tries to halve the integer, failing if it is odd.

How do we halve an integer three times?

takeOneEighth :: Int -> Maybe Int            -- (after you read the 'do' sub-section:)
takeOneEighth x =                
  case halve x of                               --  do {
    Nothing -> Nothing
    Just oneHalf ->                             --     oneHalf    <- halve x
      case halve oneHalf of
        Nothing -> Nothing
        Just oneQuarter ->                      --     oneQuarter <- halve oneHalf
          case halve oneQuarter of
            Nothing -> Nothing                  --     oneEighth  <- halve oneQuarter
            Just oneEighth ->                         
              Just oneEighth                    --     return oneEighth }

• takeOneEighth is a sequence of three halve steps chained together.
• If a halve step fails, we want the whole composition takeOneEighth to fail.
• If a halve step succeeds, we want to pipe its result forward.

instance Monad Maybe where
  -- (>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b
  Nothing >>= f  = Nothing                            -- infixl 1 >>=
  Just x  >>= f  = Just (f x)                         -- also, f =<< m = m >>= f
  
  -- return :: a -> Maybe a
  return x       = Just x

and now we can write:

takeOneEighth :: Int -> Maybe Int
takeOneEighth x = halve x >>= halve >>= halve             -- or,
    -- return x >>= halve >>= halve >>= halve             -- which is parsed as
    -- (((return x) >>= halve) >>= halve) >>= halve       -- which can also be written as
    -- (halve =<<) . (halve =<<) . (halve =<<) $ return x    -- or, equivalently, as
    --  halve <=<     halve <=<     halve      $        x

Kleisli composition <=< is defined as (g <=< f) x = g =<< f x, or equivalently as (f >=> g) x = f x >>= g. With it the above definition becomes just

takeOneEighth :: Int -> Maybe Int
takeOneEighth = halve <=< halve <=< halve               -- infixr 1 <=<
        -- or, equivalently,                    
        --      halve >=> halve >=> halve               -- infixr 1 >=>    

There are three monad laws that should be obeyed by every monad, that is every type which is an instance of the Monad typeclass:

1.  return x >>= f  =  f x
2.    m >>= return  =  m
3. (m >>= g) >>= h  =  m >>= (\y -> g y >>= h)

where m is a monad, f has type a -> m b and g has type b -> m c.

Or equivalently, using the >=> Kleisli composition operator defined above:

1.    return >=> g  =  g                    -- do { y <- return x ; g y } == g x
2.    f >=> return  =  f                    -- do { y <- f x ; return y } == f x
3. (f >=> g) >=> h  =  f >=> (g >=> h)      -- do { z <- do { y <- f x; g y } ; h z }
                                            --  == do { y <- f x ; do { z <- g y; h z } }

Obeying these laws makes it a lot easier to reason about the monad, because it guarantees that using monadic functions and composing them behaves in a reasonable way, similar to other monads.

Let's check if the Maybe monad obeys the three monad laws.

1. The left identity law - return x >>= f = f x

return z >>= f 
= (Just z) >>= f 
= f z

1. The right identity law - m >>= return = m

• Just data constructor

Just z >>= return
= return z
= Just z  

• Nothing data constructor

Nothing >>= return
= Nothing 

1. The associativity law - (m >>= f) >>= g = m >>= (\x -> f x >>= g)

• Just data constructor

-- Left-hand side
((Just z) >>= f) >>= g
= f z >>= g

-- Right-hand side
(Just z) >>= (\x -> f x >>= g)
(\x -> f x >>= g) z
= f z >>= g

• Nothing data constructor

-- Left-hand side
(Nothing >>= f) >>= g
= Nothing >>= g
= Nothing

-- Right-hand side
Nothing >>= (\x -> f x >>= g)
= Nothing

There is no way to get a value of type a out of an expression of type IO a and there shouldn't be. This is actually a large part of why monads are used to model IO.

An expression of type IO a can be thought of as representing an action that can interact with the real world and, if executed, would result in something of type a. For example, the function getLine :: IO String from the prelude doesn't mean that underneath getLine there is some specific string that I can extract - it means that getLine represents the action of getting a line from standard input.

Not surprisingly, main :: IO () since a Haskell program does represent a computation/action that interacts with the real world.

The things you can do to expressions of type IO a because IO is a monad:

  -- print the lines "Hello" then "World" to stdout
  putStrLn "Hello" >> putStrLn "World"

 -- get a line from stdin and print it back out
 getLine >>= putStrLn

 -- make an action that just returns 5
 return 5

More from the Haskell Wiki on the IO monad here (opens new window).

The lists form a monad. They have a monad instantiation equivalent to this one:

sumnd xs ys = do
  x <- xs
  y <- ys
  return (x + y)

Or equivalently, using liftM2 in Control.Monad:

sumnd = liftM2 (+)

we obtain:

> sumnd [1,2,3] [0,10]
[1,11,2,12,3,13]

do-notation is syntactic sugar for monads. Here are the rules:

example :: IO Integer
example =
  putStrLn "What's your name?" >> (
    getLine >>= (\name ->
      putStrLn ("Hello, " ++ name ++ ".") >> (
        putStrLn "What should we return?" >> (
          getLine >>= (\line ->
            let n = (read line :: Integer) in
              return (n + n))))))

example :: IO Integer
example = do
  putStrLn "What's your name?"
  name <- getLine
  putStrLn ("Hello, " ++ name ++ ".")
  putStrLn "What should we return?"
  line <- getLine
  let n = (read line :: Integer)
  return (n + n)