You can create a unbound channel with (chan), but not if you use a
transducer; (chan nil (filter odd?)) will raise an error that no buffer is
provided. Why is this the case?
Why the enforcement of all channels to be bound? In a program, there will
be channels that propagate to other channels, s
"The best code is never written"
https://zeppelin.apache.org/
https://nifi.apache.org/
Thad
https://www.linkedin.com/in/thadguidry/
On Tue, Jul 2, 2019 at 11:07 AM orazio wrote:
> Hi All,
>
> I'm newbie on Clojure/Big Data, and i'm starting with hadoop.
> I have installed Hortonworks HDP 3.1
OK that makes more sense, BeamFn is not an interface. I imagine they argued
about that decision at some point.
In your example, MyFn looks to be a general class as it can work with any
clojure var. Do you have a package with many of these type of stub classes
defined? I mean, you could name it Bea
(chan) is not a channel with an unbounded buffer. It is a channel with *no*
buffer and needs to rendezvous putters and takers 1-to-1. (Additionally it
will throw an exception if more than 1024 takers or putters are enqueued
waiting)
On Wednesday, July 3, 2019 at 7:14:46 AM UTC-4, Ernesto Garci
Hi everyone,
I'm pleased to announce the release of Deeto 0.1.0 [1]
Deeto is a Clojure library for Java developers. With Deeto you can
define your data transfer object types via interfaces in Java. You do
not need to implement these interfaces. Instead you can ask Deeto to
analyze (via reflection
Sounds like everything I escaped from when I discovered Clojure.
gvim
On 03/07/2019 22:38, henrik42 wrote:
Hi everyone,
I'm pleased to announce the release of Deeto 0.1.0 [1]
Deeto is a Clojure library for Java developers. With Deeto you can
define your data transfer object types via interfac
