On Wednesday 19 August 2009 12:14:24 am Jason McCarty wrote:
Interestingly, foldM can also be written as a left fold. To see this, note
that it is a theorem that foldr f z xs = foldl f z xs as long as f is
associative and z is a unit for f.
It must also be the case that xs is finite in length,
2009/8/19 Dan Doel dan.d...@gmail.com:
On Wednesday 19 August 2009 12:14:24 am Jason McCarty wrote:
Interestingly, foldM can also be written as a left fold. To see this, note
that it is a theorem that foldr f z xs = foldl f z xs as long as f is
associative and z is a unit for f.
This is not
Am Mittwoch 19 August 2009 16:32:57 schrieb Eugene Kirpichov:
2009/8/19 Dan Doel dan.d...@gmail.com:
On Wednesday 19 August 2009 12:14:24 am Jason McCarty wrote:
Interestingly, foldM can also be written as a left fold. To see this,
note that it is a theorem that foldr f z xs = foldl f z xs
It is associativity that is required, not commutativity (in addition to the
fact that the list is finite).
This is why Data.Foldable provides operations for monoids over containers.
Monoids just provide you with associativity and a unit, which lets you
reparenthesize the fold however you want.
You're right. My bad, indeed.
2009/8/19 Daniel Fischer daniel.is.fisc...@web.de:
Am Mittwoch 19 August 2009 16:32:57 schrieb Eugene Kirpichov:
2009/8/19 Dan Doel dan.d...@gmail.com:
On Wednesday 19 August 2009 12:14:24 am Jason McCarty wrote:
Interestingly, foldM can also be written as a
[This message is literate Haskell]
Hi -cafe,
I was trying to use Monad.foldM in the State monad recently and ran into some
performance issues. They were eventually fixed with seq, but along the way I
made some discoveries, which I thought I would share.
The Report defines foldM as
foldM
