On 2/26/07, Thomas Hartman [EMAIL PROTECTED] 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,
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]
in addition, a good example of how to apply quickcheck would be really awesome.
without using the standard drop g
On 2/26/07, Thomas Hartman [EMAIL PROTECTED] wrote:
Here's my, probably very obvious, contribution.
What I'd like feedback on is
1) code seem ok? (hope so!)
2) What do you think
ok maybe i should have read ahead. but still, i can see how to apply
hunit, but not quickcheck. but quickcheck seems more powerful.
On 2/26/07, Steve Downey [EMAIL PROTECTED] wrote:
in addition, a good example of how to apply quickcheck would be really
awesome.
without using the standard drop g
Hi I am trying to implement the function drop in haskell the thing is that I
I have been trying for some time and I came up with this code where I am
trying to do recursion:
drop :: Integer - [Integer] - [Integer]
drop 0 (x:xs) = (x:xs)
drop n (x:xs)
|n lList (x:xs) = dropN (n-1) xs :
Hey,
you're almost there:
drop :: Integer - [a] - [a]
drop 0 xs = xs
drop n (x:xs) = drop (n-1) xs
Your version fails when trying to do drop 10 [1..10]. My version
fails when trying to do drop 10 [1..9], so you might want to try to
see if you can come up with a solution for that!
Good
Hi Iliali,
You wrote:
Hi I am trying to implement the function drop in haskell
Chris Eidhof wrote:
you're almost there...
In case this is homework, you may also want to
look at:
http://www.haskell.org/haskellwiki/Homework_help
Regards,
Yitz
___
On 2/25/07, iliali16 [EMAIL PROTECTED] wrote:
Hi I am trying to implement the function drop in haskell the thing is that
I
I have been trying for some time and I came up with this code where I am
trying to do recursion:
drop :: Integer - [Integer] - [Integer]
drop 0 (x:xs) = (x:xs)
drop n
Here's my, probably very obvious, contribution.
What I'd like feedback on is
1) code seem ok? (hope so!)
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?
**
[EMAIL
