Case
Question
Can I have a case
where the alternatives contain expressions?
Answer
There are several approaches to this problem.
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')
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
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
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.