# 99 questions/Solutions/22

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

(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) |
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| start == stop = [stop] |
| start == stop = [stop] |
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| start < stop = start:range (start+1) stop |
| start < stop = start:range (start+1) stop |
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+ | </haskell> |
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+ | The following does the same but without using a reverse function |
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+ | <haskell> |
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+ | range :: Int -> Int -> [Int] |
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+ | range n m |
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+ | | n == m = [n] |
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+ | | n < m = n:(range (n+1) m) |
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+ | | n > m = n:(range (n-1) m) |
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</haskell> |
</haskell> |
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## 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.