How does the type of fromList get determined? And is Data.Set.toList the
same as Data.Set.toAscList?
Dan Piponi wrote:
If you allow me to play Devil's advocate for a moment...just don't let
this guy ask you how long the Haskell version takes.
In fact, you can borrow a trick from the C++ version. Try this instead:
import Data.Set
main = interact $ unlines . toList . fromList . words
Assuming Data.Set is implemented the way I think it is it should
perform much better than the sort . nub version.
--
Dan
On 5/18/07, Andrew Coppin <[EMAIL PROTECTED]> wrote:
OK, so I was hanging out on this newsgroup I often lurk. And we're
having a discussion. And I'm all like "Haskell is the greatest!" And
they're all like "nah, Haskell sux". And this one dude goes "hey, look
at this C++ code. How do you do that in Haskell?" And three other people
look at this C++ and can't figure out what the heck it even does. So
eventually he tells us what it does, and so I post some Haskell in a
reply. And the some other guy writes this...
(OK, not much point to this email. But it seriously made me LOL IRL!)
Invisible wrote:
>>>> int main()
>>>> {
>>>> typedef std::istream_iterator<std::string> Is;
>>>> typedef std::ostream_iterator<std::string> Os;
>>>> std::set<std::string> t((Is(std::cin)), Is());
>>>> std::copy(t.begin(), t.end(), Os(std::cout, "\n"));
>>>> }
> OK then.
> main = interact $ unlines . sort . nub . words
Pwned!
--
Darren New / San Diego, CA, USA (PST)
His kernel fu is strong.
He studied at the Shao Linux Temple.
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