Ah, OK! Another programming rule:
"When you're confused: KISS"
So instead of trying to wire a source to two transducers with tee,
I just wired a source to two sinks, so I got this to work easily then:
spawn_fthread$
src |-> xduce 42 |->
(
xduce 11 |-> sink,
xduce 22 |-> sink)
)
;
Since I only did
src |-> (sink,sink)
why does this work?
The trivial (evil) answer is the famous Skaller Cheating Method (TM).
The example works because
src -> xduce 42
is a source (using existing pipe operator) and
xduce j |-> sink
is a sink (using existing pipe operator).
So why is this a cheat? Well its a cheat because there aren't enough
overloads for it to be associative. For example this won't work:
xduce 42 |->
(
xduce 11 |-> sink,
xduce 22 |-> sink)
)
because I cannot tee a transducer into two sinks.
Similarly I cannot do:
src |-> xduce 42 |->
(
xduce 11 ,
xduce 22)
)
to get two sources.
--
john skaller
skal...@users.sourceforge.net
http://felix-lang.org
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