# Lazy pattern match

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## Latest revision as of 14:51, 6 March 2013

What does "lazy pattern match" mean and what is the meaning of the tilde in pattern matches?

## [edit] 1 Syntax

These are all lazy pattern matches:

let (a,b) = p

f ~(a,b) = ...

case p of ~(a,b) -> ...

(\ ~(a,b) -> ... )

This seems to be quite arbitrary but this is how it is defined. That is, if you want to match constructors lazily in two levels then you have to write:

let (a, ~(b,c)) = p

f ~(a, ~(b,c)) = ...

case p of ~(a, ~(b,c)) -> ...

(\ ~(a, ~(b,c)) -> ... )

## [edit] 2 Meaning

What is the meaning of a lazy pattern match and why is it required sometimes?

The lazy pattern match on a pair as in

f ~(a,b) = g a b

can be translated to

f p = g (fst p) (snd p)

Generally, a lazy pattern match is translated to calling corresponding record field accessors. The key difference between strict pattern match

f (a,b) = g a b

and lazy pattern match

f ~(a,b) = g a b

import Prelude hiding (splitAt) splitAt :: Int -> [a] -> ([a], [a]) splitAt n xs = if n<=0 then ([], xs) else case xs of [] -> ([], []) y:ys -> case splitAt (n-1) ys of ~(prefix, suffix) -> (y : prefix, suffix)

Now try

Test> splitAt 1000000 $ repeat 'a'

## [edit] 3 Implications

The lazy pattern match has some consequences. First of all a lazy pattern matches immediately always. Remember,

f ~(x:xs) = x:xs

is translated to

f ys = head ys : tail ys

f :: [a] -> [a] f [] = [] f ~(x:xs) = x:xs

is fine but stupid, because the first match already requires the decision whether the list is empty or not. But the reversed order

f :: [a] -> [a] f ~(x:xs) = x:xs f [] = []