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m (This edit provides another method of computing the range without using reverse)
(added one solution with scanl)
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 with scanl
  +
<haskell>
  +
range l r = scanl (+) l (replicate (l - r) 1)
 
</haskell>
 
</haskell>
   

Revision as of 23:15, 13 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 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.