# 99 questions/Solutions/22

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

(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.