So, I'm doing something like this
foldl (=) someA list :: Monad m = m a
where
list :: Monad m = [a - m a],
someA :: Monad m = m a
Is there a more concise way to write this? I don't think foldM is what I
want -- or is it?
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You could do:
runKleisli . mconcat . map Kleisli :: Monoid (Kleisli m a b) = [a - m b]
- a - m b
Would that work for you?
On Tue, Apr 16, 2013 at 8:35 PM, Christopher Howard
christopher.how...@frigidcode.com wrote:
So, I'm doing something like this
foldl (=) someA list :: Monad m = m a
On 04/16/2013 01:47 PM, Lyndon Maydwell wrote:
You could do:
runKleisli . mconcat . map Kleisli :: Monoid (Kleisli m a b) = [a -
m b] - a - m b
Would that work for you?
I can't find an instance for Monoid (Kleisli m a b) in `base`, so
presumably the author would also have to write this
Wow looks like this Monoid instance isn't included in Control.Monad... My
mistake.
On Tue, Apr 16, 2013 at 8:47 PM, Lyndon Maydwell maydw...@gmail.com wrote:
You could do:
runKleisli . mconcat . map Kleisli :: Monoid (Kleisli m a b) = [a - m b]
- a - m b
Would that work for you?
On
* Christopher Howard christopher.how...@frigidcode.com [2013-04-16
04:35:39-0800]
So, I'm doing something like this
foldl (=) someA list :: Monad m = m a
where
list :: Monad m = [a - m a],
someA :: Monad m = m a
Is there a more concise way to write this? I don't think foldM is
On Tue, Apr 16, 2013 at 01:53:19PM +0100, Oliver Charles wrote:
On 04/16/2013 01:47 PM, Lyndon Maydwell wrote:
You could do:
runKleisli . mconcat . map Kleisli :: Monoid (Kleisli m a b) = [a
- m b] - a - m b
Would that work for you?
I can't find an instance for Monoid (Kleisli m a b) in
Yep. I was backstabbed by ghci seemingly having no issue with my definition
when I asked for the type.
On Tue, Apr 16, 2013 at 9:49 PM, Tom Ellis
tom-lists-haskell-cafe-2...@jaguarpaw.co.uk wrote:
On Tue, Apr 16, 2013 at 01:53:19PM +0100, Oliver Charles wrote:
On 04/16/2013 01:47 PM, Lyndon
Right. See also this discussion:
http://www.haskell.org/pipermail/libraries/2009-July/012106.html
Roman
* Tom Ellis tom-lists-haskell-cafe-2...@jaguarpaw.co.uk [2013-04-16
14:49:48+0100]
On Tue, Apr 16, 2013 at 01:53:19PM +0100, Oliver Charles wrote:
On 04/16/2013 01:47 PM, Lyndon Maydwell
Hi!
Although foldM won't make things much nicer, it can be used here as well:
someA = \a - foldM (flip id) a list
Cheers!
Arseniy
On 16 April 2013 13:35, Christopher Howard
christopher.how...@frigidcode.com wrote:
So, I'm doing something like this
foldl (=) someA list :: Monad m = m a
