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Hand off control between java threads


Hand off control between java threads

By : João Almeida
Date : November 29 2020, 09:01 AM
should help you out The simplest way to implement this is just to not use threads in the first place. You can still separate the code in different methods and classes.
Assuming you need to use threads for some reason then this can be done with a simple Object lock = new Object() along with wait and notifyAll.
code :


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Can you control the latency for java threads?

Can you control the latency for java threads?


By : paddy_mack
Date : March 29 2020, 07:55 AM
this one helps. No you can't.
What you have here is 2 absolutely different activities coupled together. One is responsible for periodical wakeups and the other takes care of updating the client. If the wakeup task waits for update task to be done then its accuracy will suffer.
code :
ScheduledExecutorService executor = Executors.newScheduledThreadPool(1);

    final Runnable notificationTask = new Runnable() {
        @Override public void run() {
            updater.sendUpdate();                          // just *notify* the updater to do an update
        }
    };

    executor.scheduleAtFixedRate(
        notificationTask, 0, 33333333, TimeUnit.NANOSECONDS // freakishly accurate :)
    );
JAVA real time consol-control the threads

JAVA real time consol-control the threads


By : pawan kumar
Date : March 29 2020, 07:55 AM
This might help you I am a beginner in programming. I am still learning about threads and more of them. But now I have quite a big idea to write my first program (I mean bigger than simple calculator). I want it to sort files, integrate in one (many copies of one file in different localization - the idea of it is of no importance now). , A simple way to do it using the standard library :
code :
import java.util.Scanner;
import java.util.concurrent.LinkedBlockingDeque;
import java.util.concurrent.ThreadPoolExecutor;

import static java.util.concurrent.TimeUnit.MILLISECONDS;

public class Example {
    private static final int POOL_SIZE = 5;
    private static final ExecutorService WORKERS = new ThreadPoolExecutor(POOL_SIZE, POOL_SIZE, 1, MILLISECONDS, new LinkedBlockingDeque<>());

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (true) {
            System.out.print("> ");
            String cmd = sc.nextLine();
            switch (cmd) {
                case "process":
                    WORKERS.submit(newExpensiveTask());
                    break;

                case "kill":
                    System.exit(0);

                default:
                    System.err.println("Unrecognized command: " + cmd);
            }
        }
    }

    private static Runnable newExpensiveTask() {
        return () -> {
            try {
                Thread.sleep(10000);
                System.out.println("Done processing");
            } catch (InterruptedException e) {
                throw new RuntimeException(e);
            }
        };
    }
}
Java locks: Hand over hand locking through list

Java locks: Hand over hand locking through list


By : Michael Ladiona
Date : March 29 2020, 07:55 AM
this will help Looks like maybe you don't understand what a Lock is.
A Lock object can be in either be owned (a.k.a., locked) by some thread, or it can be available (a.k.a., unlocked).
Solving deadlock in hand-over-hand locking a Linked list in Java

Solving deadlock in hand-over-hand locking a Linked list in Java


By : Grushbrush
Date : March 29 2020, 07:55 AM
wish help you to fix your issue In the while loop inside add(), you actually have a couple of problems.
Your pred node doesn't get unlocked. You lock the curr node twice.
Determine if an Indian Rummy hand is a winning hand - Java

Determine if an Indian Rummy hand is a winning hand - Java


By : ゴミゴキブリチャンネラ
Date : March 29 2020, 07:55 AM
help you fix your problem I have implemented a java version of Rummikub (a game with similar constraints).
My approach there was to give each card a hidden integer attribute (a prime number).
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