Factory function

Categories: Idioms

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 1.2 Red-black trees example 2 See also

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.

User:AndrewBromage

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:

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

To do this, we create a factory function, balance that ensures the invariants are met.

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)

2 See also

• Hash consing
• Memoising constant applicative forms.

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