Personal tools

99 questions/Solutions/22

From HaskellWiki

< 99 questions | Solutions(Difference between revisions)
Jump to: navigation, search
(slightly modified the version with guards so that it will make backwards ranges too)
m (This edit provides another method of computing the range without using reverse)
Line 18: Line 18:
 
| start == stop = [stop]
 
| start == stop = [stop]
 
| start < stop = start:range (start+1) stop
 
| start < stop = start:range (start+1) stop
  +
</haskell>
  +
The following does the same but without using a reverse function
  +
<haskell>
  +
range :: Int -> Int -> [Int]
  +
range n m
  +
| n == m = [n]
  +
| n < m = n:(range (n+1) m)
  +
| n > m = n:(range (n-1) m)
 
</haskell>
 
</haskell>
   

Revision as of 01:23, 17 March 2011

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)

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.