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
in addition, a good example of how to apply quickcheck would be really awesome.
without using the standard drop
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
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
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,
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
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
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
___
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 l
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 :