# Case

### From HaskellWiki

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Alternatively, one could simply factor out a function(/value) and use guards in the argument patterns. |
Alternatively, one could simply factor out a function(/value) and use guards in the argument patterns. |
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+ | === Using list comprehensions === |
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+ | |||

+ | An alternative sugarful approach is to use [[list comprehension]]s. |
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+ | |||

+ | <haskell> |
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+ | head $ |
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+ | [ ex1 | cond1 ] ++ |
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+ | [ ex2 | cond2 ] ++ |
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+ | [ ex3 | cond3 ] ++ |
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+ | [ exDefault ] |
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+ | </haskell> |
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[[Category:FAQ]] |
[[Category:FAQ]] |

## Revision as of 04:36, 29 January 2008

## Contents |

## 1 Question

Can I have acase

## 2 Answer

There are several approaches to this problem.

### 2.1 Using functions

We can do this nicely with a function implemented in Haskell:

select :: a -> [(Bool, a)] -> a select def = maybe def snd . List.find fst select exDefault [(cond1, ex1), (cond2, ex2), (cond3, ex3)]

Unfortunately this function is not in the Prelude.

Alternative implementations are

select' def = fromMaybe def . lookup True {- a purely functional implementation of if-then-else -} if' :: Bool -> a -> a -> a if' True x _ = x if' False _ y = y select'' = foldr (uncurry if')

select''

select

if

if'

zipWith3

zipWith3 if'

See if-then-else.

Alternatively you can unrollfoldr

if' cond1 ex1 $ if' cond2 ex2 $ if' cond3 ex3 $ exDefault

if'

?

then because of partial application it will work nicely together with '$' for the else clause.

infixl 1 ? (?) :: Bool -> a -> a -> a (?) = if' cond1 ? ex1 $ cond2 ? ex2 $ cond3 ? ex3 $ exDefault

### 2.2 Using syntactic sugar

You can make use of some syntactic sugar of Haskell, namely of guards.

case () of _ | cond1 -> ex1 | cond2 -> ex2 | cond3 -> ex3 | otherwise -> exDefault

Alternatively, one could simply factor out a function(/value) and use guards in the argument patterns.

### 2.3 Using list comprehensions

An alternative sugarful approach is to use list comprehensions.

head $ [ ex1 | cond1 ] ++ [ ex2 | cond2 ] ++ [ ex3 | cond3 ] ++ [ exDefault ]