Re: [Haskell-cafe] Hi can u explain me how drop works in Haskell

I'd heard of quick check, but haven't got my head around it. This seems like a good place to start. I understand you have to build an invariant and then you can automate against it, eg "reverse of reverse is your original string" prop_RevRev xs = reverse (reverse xs) == xs where types = xs::[Int] (from http://www.kimbly.com/blog/000042.html) but what would be your invariant with the drop function? At first I thought prop_HeadConsDropped xs = (head xs) : (drop 1 xs) == xs where types = xs::[a] But I think then you have the issue of head nil being an error. So, is there a better strategy? Or is using quickcheck here overkill? 2007/2/26, Antonio Cangiano :

On 2/26/07, Thomas Hartman wrote:

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Here's my, probably very obvious, contribution.

What I'd like feedback on is

1) code seem ok? (hope so!)

Hi Thomas,

tail [] raises an error, therefore your code will fail when n > length xs ( e.g. mydrop 3 [1,2] will raise an exception, where [] is the expected result). Your function is also limited to list of Int only (mydrop :: Int -> [Int] -> [Int]).

...

2) What do you think of the tests I did to verify that this behaves the way I want? Is there a better / more idiomatic way to do this?

You may be interested in the following projects:

QuickCheck: http://www.cs.chalmers.se/~rjmh/QuickCheck/ HUnit: http://sourceforge.net/projects/hunit

Regards, Antonio -- http://antoniocangiano.com Zen and the Art of Ruby Programming