Difference between revisions of "99 questions/Solutions/22"

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(generic version similar to one of the others)
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Since there's already syntactic sugar for ranges, there's usually no reason to define a function like 'range' in Haskell. In fact, the syntactic sugar is implemented using the enumFromTo function, which is exactly what 'range' should be.
 
Since there's already syntactic sugar for ranges, there's usually no reason to define a function like 'range' in Haskell. In fact, the syntactic sugar is implemented using the enumFromTo function, which is exactly what 'range' should be.
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[[Category:Programming exercise spoilers]]

Revision as of 19:38, 18 January 2014

Create a list containing all integers within a given range.

range x y = [x..y]

or

range = enumFromTo

or

range x y = take (y-x+1) $ iterate (+1) x

or

range start stop
    | start > stop  = reverse (range stop start)
    | start == stop = [stop]
    | start < stop  = start:range (start+1) stop

The following does the same but without using a reverse function

range :: Int -> Int -> [Int]
range n m
    | n == m = [n]
    | n < m = n:(range (n+1) m)
    | n > m = n:(range (n-1) m)

or, a generic and shorter version of the above

range :: (Ord a, Enum a) => a -> a -> [a]
range a b | (a == b) = [a]
range a b = a:range ((if a < b then succ else pred) a) b

or with scanl

range l r = scanl (+) l (replicate (l - r) 1)

Since there's already syntactic sugar for ranges, there's usually no reason to define a function like 'range' in Haskell. In fact, the syntactic sugar is implemented using the enumFromTo function, which is exactly what 'range' should be.