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

[Thomas Hartman]

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 <acangiano at gmail.com>:
> On 2/26/07, Thomas Hartman <tphyahoo at gmail.com> wrote:
> > 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