Portability | non-portable (fundeps, MPTCs) |
---|---|

Stability | experimental |

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

Safe Haskell | Safe-Inferred |

Monads for free.

# Documentation

class Monad m => MonadFree f m | m -> f whereSource

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, the `kan-extensions`

package provides `MonadFree`

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

).

See `Free`

for a more formal definition of the free `Monad`

for a `Functor`

.