Actually, presumably you meant:
> reverse2 ys (x:xs) = reverse2 (x:ys) xs
(stupid cut & paste late at night).
with this fix (and now we actually are reversing the list with reverse2),
we get the following timings for reverse2:
11:49pm enescu:~/ time a.out
200
4.64u 0.31s 0:05.18 95.5%
11:49
Well, I assume you meant:
reverse1 [] ys = ys
reverse1 (x:xs) ys = reverse1 xs (x:ys)
reverse2 ys [] = ys
reverse2 ys (x:xs) = reverse1 (x:ys) xs
If so, and you make two programs:
main = print (length $! reverse1 [1..200] [])
and
main = print (length $! reverse2 [] [1..200])
compile
Hi.
I want to join please.
thanks in advance.
Ilan.
At 2002-02-06 03:38, John Hughes wrote:
>Well, I'm still not convinced. A reference *value* can't have the type
>
> (LiftedMonad (ST s) m) => Ref m a
Oh, yeah, you're right. I made a mistake here:
newSTRef :: a -> Ref (ST s) a;
newSTLiftedRef :: (LiftedMonad
Hello.
Please, tell me which set of definitions below should I expected
to be more efficient: the reverse1 or the reverse2 functions.
reverse1 [] ys = ys
reverse1 (x:xs) ys = reverse2 (x:ys) xs
reverse2 ys [] = ys
reverse2 ys (x:xs) = reverse2 (x:ys) xs
The difference rely on the posit
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Ashley Yakeley wrote:
At 2002-02-06 01:09, John Hughes wrote:
>No no no! This still makes the reference type depend on the monad type, which
>means that I cannot manipulate the same reference in two different monads!
Yes you can. Consider:
-- m some
At 2002-02-06 00:54, Koen Claessen wrote:
>You are completely right, of course I meant r -> m!
Right. There's the equivalent of an r -> m dependency because m is a
parameter in the Ref type constructor. But it doesn't matter, because you
can create values of this type:
myIntRef :: (MyMon
At 2002-02-06 01:09, John Hughes wrote:
>No no no! This still makes the reference type depend on the monad type, which
>means that I cannot manipulate the same reference in two different monads!
Yes you can. Consider:
-- m somehow uses 'rep' internally
class (Monad rep, Monad m) => Lif
John Hughes despaired:
| Oh no, please don't do this! I use the RefMonad class,
| but *without* the dependency r -> m. Why not? Because
| I want to manipulate (for example) STRefs in monads
| built on top of the ST monad via monad transformers.
At 2002-02-06 00:33, Koen Claessen wrote:
>Hm... this looks nice. With slight name changes this
>becomes:
Oh if you must. I decided that Refs were _so_ fundamental that anytime
you get, set or modify anything it could probably be represented as a
Ref, so the functions merit highly generic name
John Hughes despaired:
| Oh no, please don't do this! I use the RefMonad class,
| but *without* the dependency r -> m. Why not? Because
| I want to manipulate (for example) STRefs in monads
| built on top of the ST monad via monad transformers.
| So I use the same reference type with *many d
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