Copyright | (C) 2008-2015 Edward Kmett |
---|---|

License | BSD-style (see the file LICENSE) |

Maintainer | Edward Kmett <ekmett@gmail.com> |

Stability | experimental |

Portability | non-portable (fundeps, MPTCs) |

Safe Haskell | Safe |

Language | Haskell2010 |

Monads for free.

# Documentation

class Monad m => MonadFree f m | m -> f where Source

Monads provide substitution (`fmap`

) and renormalization (`join`

):

m`>>=`

f =`join`

(`fmap`

f m)

A free `Monad`

is one that does no work during the normalization step beyond simply grafting the two monadic values together.

`[]`

is not a free `Monad`

(in this sense) because

smashes the lists flat.`join`

[[a]]

On the other hand, consider:

data Tree a = Bin (Tree a) (Tree a) | Tip a

instance`Monad`

Tree where`return`

= Tip Tip a`>>=`

f = f a Bin l r`>>=`

f = Bin (l`>>=`

f) (r`>>=`

f)

This `Monad`

is the free `Monad`

of Pair:

data Pair a = Pair a a

And we could make an instance of `MonadFree`

for it directly:

instance`MonadFree`

Pair Tree where`wrap`

(Pair l r) = Bin l r

Or we could choose to program with

instead of `Free`

Pair`Tree`

and thereby avoid having to define our own `Monad`

instance.

Moreover, Control.Monad.Free.Church provides a `MonadFree`

instance that can improve the *asymptotic* complexity of code that
constructs free monads by effectively reassociating the use of
(`>>=`

). You may also want to take a look at the `kan-extensions`

package (http://hackage.haskell.org/package/kan-extensions).

See `Free`

for a more formal definition of the free `Monad`

for a `Functor`

.

Nothing