"Jorge Marques Pelizzoni" <[EMAIL PROTECTED]> escribió:
Hi Julio,
Look, your example does not work because if the element is in both lists
both threads are going to try and determine A, which will necessarily
fail.
Also nobody ever binds A to some value indicating the element is not in
either list.
Here is a version of something similar that would work:
------------------------
proc {TheFirstWins Tasks Rslt Success}
Rslts
RsltPort = {NewPort Rslts}
AllFinished = {FoldL Tasks
proc {$ PrevFinished Task ImFinished}
TaskThread
in
thread
TaskThread = {Thread.this}
try
MyRslt = {Task}
in
if {Not {IsFree MyRslt}} then
{Send RsltPort MyRslt}
{Wait Rslts} % : just in case
end
catch X then
{Send RsltPort {Value.failed X}}
{Wait Rslts}
end
PrevFinished = ImFinished
{Wait _}
end
thread % thread killer:
{Wait Rslts}
{Thread.terminate TaskThread}
end
end
unit}
in
{Record.waitOr Rslts#AllFinished _}
Success = {Not {IsFree Rslts}}
if Success then Rslt = Rslts.1 else
{Send RsltPort unit} % enables thread-killing
end
end
--------------
I haven't tested it, but conceptually it must be ok. Tasks here is a list
of unary procedures, which could look like:
[ fun {$} {Esta E L1 $ 1} end
fun {$} {Esta E L2 $ 2} end
]
I also included some sort of exception handling.
Cheers,
Jorge.
[EMAIL PROTECTED] escreveu:
Hello
An example I think for my students. A concurrent search for an element
in two lists
declare
proc {Esta E L ?A Id}
case L
of nil then skip
[]X|Xs then if X==E then A=Id#true else {Esta E Xs ?A Id} end end
end
proc {Esta2 E L1 L2 ?A}
thread {Esta E L1 ?A 1} end
thread {Esta E L2 ?A 2} end
end
declare
Ls1= [1 3 5 7 9 11 13 15]
Ls2= [2 4 6 8 10 12 14 16 18]
R={Esta2 13 Ls1 Ls2} {Browse R}
It seems works. It's a good example? The question is: When one of the
threads binds ?A, the others threads continue making the search. How can
one stop the others threads to continue working?
Thanks (apologize for my english)
Julio H Vargas
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Jorge M. Pelizzoni
ICMC - Universidade de São Paulo
call me at http://www.jaxtr.com/jpeliz
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Thanks a lot. My example is so simple that I make a precondition: E
actually is in only one of the lists.
Julio Hernando
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