Case
From HaskellWiki
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=== Using functions === | === Using functions === | ||
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| + | ==== select ==== | ||
We can do this nicely with a function implemented in Haskell: | We can do this nicely with a function implemented in Haskell: | ||
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Unfortunately this function is not in the [[Prelude]]. | Unfortunately this function is not in the [[Prelude]]. | ||
It is however in the [http://hackage.haskell.org/packages/archive/utility-ht/0.0.1/doc/html/Data-Bool-HT.html#v%3Aselect utility-ht] package. | It is however in the [http://hackage.haskell.org/packages/archive/utility-ht/0.0.1/doc/html/Data-Bool-HT.html#v%3Aselect utility-ht] package. | ||
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| + | ==== nested 'if' ==== | ||
Alternative implementations are | Alternative implementations are | ||
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exDefault | exDefault | ||
</haskell> | </haskell> | ||
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| + | ==== infix operator ==== | ||
If you use <hask>if'</hask> in infix form, | If you use <hask>if'</hask> in infix form, | ||
Revision as of 10:26, 8 January 2009
Contents |
1 Question
Can I have acase
2 Answer
There are several approaches to this problem.
2.1 Using functions
2.1.1 select
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. It is however in the utility-ht package.
2.1.2 nested 'if'
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
2.1.3 infix operator
If you useif'
?
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
2.2.1 Guards
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.
2.2.2 List comprehensions
An alternative sugarful approach is to use list comprehensions.
head $ [ ex1 | cond1 ] ++ [ ex2 | cond2 ] ++ [ ex3 | cond3 ] ++ [ exDefault ]
Categories: FAQ | Idioms
