99 questions/Solutions/22: Difference between revisions
< 99 questions | Solutions
(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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| 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> | ||
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 01:23, 17 March 2011
Create a list containing all integers within a given range.
range x y = [x..y]or
range = enumFromToor
range x y = take (y-x+1) $ iterate (+1) xor
range start stop
| start > stop = reverse (range stop start)
| start == stop = [stop]
| start < stop = start:range (start+1) stopThe 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.