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99 questions/Solutions/22

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(added one solution with scanl)
(generic version similar to one of the others)
Line 26: Line 26:
 
| n < m = n:(range (n+1) m)
 
| n < m = n:(range (n+1) m)
 
| n > m = n:(range (n-1) m)
 
| n > m = n:(range (n-1) m)
  +
</haskell>
  +
or, a generic and shorter version of the above
  +
<haskell>
  +
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
 
</haskell>
 
</haskell>
 
or with scanl
 
or with scanl

Revision as of 09:11, 24 January 2012

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.