Yes, you're right about the type context. I always forget that Functor is not a superclass of Monad. Anyway, the `fmap` can be done only with `Monad` in the context. About the list fusion, I'm compiling with optimizations enabled (-O2), but I do not know how it works with fusions. Is it the optimization omitting the step of creating the list? How can I know it with certainty? On Fri, Jan 25, 2013 at 7:23 AM, Thomas Horstmeyer < horstmey@mathematik.uni-marburg.de> wrote:
Hi,
isn't the correct type context for f the following?
f :: (Functor m, Monad m) => Int -> m a -> m (Seq a)
So your f really is a kind of specialization of g.
Could the reason for f performing better be list fusion? Anything happening inside Control.Monad.replicateM should be subject to it.
Cheers, Thomas
Am 24.01.2013 15:31, schrieb Daniel Díaz Casanueva:
Hi Cafe,
I was coding this morning when I suddenly found something that surprised me. It's been a short time since I am really caring about the performance of my programs. Before, I was just caring about their correctness. So I am trying different things and profiling to see differences. One difference I have found surprising is that the function f is MUCH faster and less space consuming than the function g:
import Control.Monad import qualified Data.Sequence as Seq
type Seq = Seq.Seq
f :: Monad m => Int -> m a -> m (Seq a) f n = fmap Seq.fromList . replicateM n
g :: Monad m => Int -> m a -> m (Seq a) g = Seq.replicateM
Maybe is just in my test case, where the Int argument is big and the monadic action short, but it looks to me that Data.Sequence.replicateM can be faster than it is right now.
Regards, Daniel Díaz.
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let f x = x in x
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