Copyright | (C) 2012-16 Edward Kmett |
---|---|

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

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

Stability | provisional |

Portability | Rank2Types |

Safe Haskell | None |

Language | Haskell2010 |

One commonly asked question is: can we combine two lenses,

and `Lens'`

a b

into `Lens'`

a c

.
This is fair thing to ask, but such operation is unsound in general.
See `Lens'`

a (b, c)`lensProduct`

.

## Synopsis

- lensProduct :: ALens' s a -> ALens' s b -> Lens' s (a, b)
- prismSum :: APrism s t a b -> APrism s t c d -> Prism s t (Either a c) (Either b d)
- adjoin :: Traversal' s a -> Traversal' s a -> Traversal' s a

# Documentation

lensProduct :: ALens' s a -> ALens' s b -> Lens' s (a, b) Source #

A lens product. There is no law-abiding way to do this in general.
Result is only a valid `ReifiedLens`

if the input lenses project disjoint parts of
the structure `s`

. Otherwise "you get what you put in" law

`view`

l (`set`

l v s) ≡ v

is violated by

`>>>`

`let badLens :: Lens' (Int, Char) (Int, Int); badLens = lensProduct _1 _1`

`>>>`

(2,2)`view badLens (set badLens (1,2) (3,'x'))`

but we should get `(1,2)`

.

Are you looking for `alongside`

?

prismSum :: APrism s t a b -> APrism s t c d -> Prism s t (Either a c) (Either b d) Source #

A dual of `lensProduct`

: a prism sum.

The law

`preview`

l (`review`

l b) ≡`Just`

b

breaks with

`>>>`

`let badPrism :: Prism' (Maybe Char) (Either Char Char); badPrism = prismSum _Just _Just`

`>>>`

Just (Left 'x')`preview badPrism (review badPrism (Right 'x'))`

We put in `Right`

value, but get back `Left`

.

Are you looking for `without`

?

adjoin :: Traversal' s a -> Traversal' s a -> Traversal' s a Source #