# 99 questions/Solutions/22

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< 99 questions | Solutions(Difference between revisions)

(slightly modified the version with guards so that it will make backwards ranges too) |
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Line 14: | Line 14: | ||

or | or | ||

<haskell> | <haskell> | ||

− | range | + | range start stop |

− | | | + | | start > stop = reverse (range stop start) |

− | | | + | | start == stop = [stop] |

+ | | start < stop = start:range (start+1) stop | ||

</haskell> | </haskell> | ||

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. | 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. |

## Revision as of 05:22, 20 November 2010

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

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.