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Can I have a {{{case}}} where the alternatives contain expressions?
+
== Question ==
   
  +
Can I have a <hask>case</hask> where the alternatives contain expressions?
   
You can make use of some SyntacticSugar of Haskell, namely of ["Guards"].
+
== Answer ==
   
{{{#!syntax haskell
+
There are several approaches to this problem.
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.
+
=== Using functions ===
   
Why sticking to syntactic sugar? We can do it nicely with a function implemented in Haskell:
+
==== select ====
{{{#!syntax haskell
+
  +
We can do this nicely with a function implemented in Haskell:
  +
<haskell>
 
select :: a -> [(Bool, a)] -> a
 
select :: a -> [(Bool, a)] -> a
 
select def = maybe def snd . List.find fst
 
select def = maybe def snd . List.find fst
+
-- = fromMaybe def . lookup True
  +
-- = maybe def id . lookup True
   
 
select exDefault
 
select exDefault
Line 18: Line 19:
 
(cond2, ex2),
 
(cond2, ex2),
 
(cond3, ex3)]
 
(cond3, ex3)]
}}}
+
</haskell>
  +
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.
   
Alternative implementations are
+
==== nested 'if' ====
{{{#!syntax haskell
 
select' def = fromMaybe def . lookup True
 
   
  +
Alternative implementations are
  +
<haskell>
 
{- a purely functional implementation of if-then-else -}
 
{- a purely functional implementation of if-then-else -}
 
if' :: Bool -> a -> a -> a
 
if' :: Bool -> a -> a -> a
Line 28: Line 31:
   
 
select'' = foldr (uncurry if')
 
select'' = foldr (uncurry if')
}}}
+
</haskell>
The implementation of {{{select''}}} makes clear that {{{select}}} can be considered as nested {{{if}}}s.
+
The implementation of <hask>select''</hask> makes clear that <hask>select</hask> can be considered as nested <hask>if</hask>s.
The functional {{{if'}}} is also useful in connection with {{{zipWith3}}} since {{{zipWith3 if'}}} merges two lists according to a list of conditions.
+
The functional <hask>if'</hask> is also useful in connection with <hask>zipWith3</hask> since <hask>zipWith3 if'</hask> merges two lists according to a list of conditions.
  +
See [[if-then-else]].
   
  +
Alternatively you can unroll <hask>foldr</hask> and write
  +
<haskell>
  +
if' cond1 ex1 $
  +
if' cond2 ex2 $
  +
if' cond3 ex3 $
  +
exDefault
  +
</haskell>
   
If you don't like the parentheses for the pairs, you can also define
+
==== infix operator ====
{{{#!syntax haskell
 
data SelectBranch a = (:->) {
 
condition :: Bool,
 
expression :: a
 
}
 
   
select :: a -> [SelectBranch a] -> a
+
If you use <hask>if'</hask> in infix form,
select def = maybe def expression . List.find condition
+
you may call it <hask>?</hask> like in C,
  +
then because of partial application it will work nicely together with '$' for the else clause.
  +
<haskell>
  +
infixl 1 ?
  +
(?) :: Bool -> a -> a -> a
  +
(?) = if'
   
  +
cond1 ? ex1 $
  +
cond2 ? ex2 $
  +
cond3 ? ex3 $
  +
exDefault
  +
</haskell>
   
select exDefault
+
=== Using syntactic sugar ===
[cond1 :-> ex1,
+
cond2 :-> ex2,
+
==== Guards ====
cond3 :-> ex3]
+
}}}
+
You can make use of some [[syntactic sugar]] of Haskell, namely of [[guard]]s.
  +
  +
<haskell>
  +
case () of _
  +
| cond1 -> ex1
  +
| cond2 -> ex2
  +
| cond3 -> ex3
  +
| otherwise -> exDefault
  +
</haskell>
  +
  +
Alternatively, one could simply factor out a function(/value) and use guards in the argument patterns.
  +
  +
==== List comprehensions ====
  +
  +
An alternative sugarful approach is to use [[list comprehension]]s.
  +
  +
<haskell>
  +
head $
  +
[ ex1 | cond1 ] ++
  +
[ ex2 | cond2 ] ++
  +
[ ex3 | cond3 ] ++
  +
[ exDefault ]
  +
</haskell>
  +
  +
[[Category:FAQ]]
  +
[[Category:Idioms]]

Latest revision as of 18:31, 15 March 2012

Contents

[edit] 1 Question

Can I have a
case
where the alternatives contain expressions?

[edit] 2 Answer

There are several approaches to this problem.

[edit] 2.1 Using functions

[edit] 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
        -- = fromMaybe def . lookup True
        -- = maybe def id . lookup True
 
select exDefault
    [(cond1, ex1),
     (cond2, ex2),
     (cond3, ex3)]

Unfortunately this function is not in the Prelude. It is however in the utility-ht package.

[edit] 2.1.2 nested 'if'

Alternative implementations are

{- 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')
The implementation of
select''
makes clear that
select
can be considered as nested
if
s. The functional
if'
is also useful in connection with
zipWith3
since
zipWith3 if'
merges two lists according to a list of conditions.

See if-then-else.

Alternatively you can unroll
foldr
and write
if' cond1 ex1 $
if' cond2 ex2 $
if' cond3 ex3 $
   exDefault

[edit] 2.1.3 infix operator

If you use
if'
in infix form, you may call it
?
like in C,

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

[edit] 2.2 Using syntactic sugar

[edit] 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.

[edit] 2.2.2 List comprehensions

An alternative sugarful approach is to use list comprehensions.

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