# Factory function

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data Expr = EAdd Expr Expr | EMult Expr Expr | EInt Int | EVar String | data Expr = EAdd Expr Expr | EMult Expr Expr | EInt Int | EVar String | ||

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Keeping an expression in a relatively simplified form can be difficult if it is modified a lot. One simple way is to write replacements for the constructor functions: | Keeping an expression in a relatively simplified form can be difficult if it is modified a lot. One simple way is to write replacements for the constructor functions: | ||

<haskell> | <haskell> |

## Revision as of 22:44, 2 December 2006

If you need more intelligence from your constructor functions, use a real function instead. Also known as smart constructors.

## Contents |

## 1 Examples

### 1.1 Expression type

Consider the following data type:

data Expr = EAdd Expr Expr | EMult Expr Expr | EInt Int | EVar String

Keeping an expression in a relatively simplified form can be difficult if it is modified a lot. One simple way is to write replacements for the constructor functions:

eInt i = EInt i eAdd (EInt i1) (EInt i2) = eInt (i1+i2) eAdd (EInt 0) e2 = e2 eAdd e1 (EInt 0) = e1 eAdd e1 e2 = EAdd e1 e2 eMult (EInt 0) e2 = eInt 0 {- etc -}

Then if you need to construct an expression, use the factory functions:

derivative :: String -> Expr -> Expr derivative x (EMult e1 e2) = eAdd (eMult (derivative x e1) e2) (eMult e1 (derivative x e2)) {- etc -}

This is actually a special kind of worker wrapper where the wrapper does more work than the worker.

The factory function idiom is especially useful when you have a data structure with invariants that you need to preserve, such as a binary search tree which needs to stay balanced.

### 1.2 Red-black trees example

This form of balanced tree is a perfect example of the use of this idiom. The type declaration for a Red-Black tree is:

data Colour = R | B deriving (Eq, Show, Ord) data RBSet a = Empty | RBTip Colour (RBSet a) a (RBSet a) deriving Show

However, this must satisfy these invariants:

- The children of a red node are black.
- There are the same number of black nodes on every path from root to leaf.

balance :: Colour -> RBSet a -> a -> RBSet a -> RBSet a balance B (RBTip R (RBTip R a x b) y c) z d = RBTip R (RBTip B a x b) y (RBTip B c z d) balance B (RBTip R a x (RBTip R b y c)) z d = RBTip R (RBTip B a x b) y (RBTip B c z d) balance B a x (RBTip R (RBTip R b y c) z d) = RBTip R (RBTip B a x b) y (RBTip B c z d) balance B a x (RBTip R b y (RBTip R c z d)) = RBTip R (RBTip B a x b) y (RBTip B c z d) balance c a x b = RBTip c a x b

(See Red-black trees in a functional setting by Chris Okasaki)