Difference between revisions of "Monad"

From HaskellWiki
Jump to navigation Jump to search
(fmt)
Line 2: Line 2:
 
The '''Monad''' class is defined like this:
 
The '''Monad''' class is defined like this:
   
  +
<haskell>
class Monad m where
+
class Monad m where
(>>=) :: m a -> (a -> m b) -> m b
 
(>>) :: m a -> m b -> m b
+
(>>=) :: m a -> (a -> m b) -> m b
return :: a -> m a
+
(>>) :: m a -> m b -> m b
fail :: String -> m a
+
return :: a -> m a
  +
fail :: String -> m a
  +
</haskell>
   
 
All instances of Monad should obey:
 
All instances of Monad should obey:
   
  +
<haskell>
return a >>= k = k a
 
m >>= return = m
+
return a >>= k = k a
m >>= (\x -> k x >>= h) = (m >>= k) >>= h
+
m >>= return = m
  +
m >>= (\x -> k x >>= h) = (m >>= k) >>= h
  +
</haskell>
   
 
Any Monad can be made a [[Functor]] by defining
 
Any Monad can be made a [[Functor]] by defining
   
  +
<haskell>
fmap ab ma = ma >>= (return . ab)
+
fmap ab ma = ma >>= (return . ab)
  +
</haskell>
   
 
However, the Functor class is not a superclass of the Monad class. See [[Functor hierarchy proposal]].
 
However, the Functor class is not a superclass of the Monad class. See [[Functor hierarchy proposal]].

Revision as of 23:27, 8 March 2006

Monad class (base)
import Control.Monad

The Monad class is defined like this:

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

All instances of Monad should obey:

return a >>= k  =  k a
m >>= return  =  m
m >>= (\x -> k x >>= h)  =  (m >>= k) >>= h

Any Monad can be made a Functor by defining

fmap ab ma = ma >>= (return . ab)

However, the Functor class is not a superclass of the Monad class. See Functor hierarchy proposal.

Monad Tutorials

Monads are known for being deeply confusing to lots of people, so there are plenty of tutorials specifically related to monads. Each takes a different approach to Monads, and hopefully everyone will find something useful.