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(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>
  
 
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 = 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.