I am trying to learn multi-threads, and parallel execution in Java. I wrote example code like this:
public class MemoryManagement1 {
public static int counter1 = 0;
public static int counter2 = 0;
public static final Object lock1= new Object();
public static final Object lock2= new Object();
public static void increment1() {
synchronized(lock1) {
counter1 ++;
}
}
public static void increment2() {
synchronized(lock2) {
counter2 ++;
}
}
public static void processes() {
Thread thread1 = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < 4; i++) {
increment1();
}
}
});
Thread thread2 = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < 4; i++) {
increment2();
}
}
});
thread1.start();
thread2.start();
try {
thread1.join();
thread2.join();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("Counter value is :" + counter1);
System.out.println("Counter value is :" + counter2);
}
public static void main(String[] args) {
processes();
}
}
The code is running properly, but how can I know that the code is running according to time-slicing or whether it is running with parallel execution. I have a CPU with 4 cores. As I understand it, the program should be run with parallel execution, but I am not sure.
The code is running properly, but how can I know that the code is
running according to time-slicing or whether it is running with
parallel execution.
A complete answer to this question would have to cover several factors, but I will be concise and focus mainly on the two most relevant points (IMO) to this question. For simplicity, let us assume that whenever possible each thread (created by the application) will be assigned to a different core.
First, it depends on the number of cores of the hardware that the application is being executed on, and how many threads (created by the application) are running simultaneously. For instance, if the hardware only has a single core or if the application creates more threads than the number of cores available, then some of those threads will inevitably not be executing truly in parallel (i.e., will be mapped to the same core).
Second, it depends if the threads executing their work synchronize with each other or not. In your code, two threads are created, synchronizing using a different object, and since your machine has 4 cores, in theory, each thread is running in parallel to each other.
It gets more complex than that because you can have parts of your code that are executed in parallel, and other parts that are executed sequentially by the threads involved. For instance, if the increment1 and increment2 methods were synchronizing on the same object, then those methods would not be executed in parallel.
Your program is indeed running in parallel execution. In this particular example however you don't need locks in your code, it would run perfectly well without them.
Related
I want the final count to be 10000 always but even though I have used synchronized here, Im getting different values other than 1000. Java concurrency newbie.
public class test1 {
static int count = 0;
public static void main(String[] args) throws InterruptedException {
int numThreads = 10;
Thread[] threads = new Thread[numThreads];
for(int i=0;i<numThreads;i++){
threads[i] = new Thread(new Runnable() {
#Override
public void run() {
synchronized (this) {
for (int i = 0; i < 1000; i++) {
count++;
}
}
}
});
}
for(int i=0;i<numThreads;i++){
threads[i].start();
}
for (int i=0;i<numThreads;i++)
threads[i].join();
System.out.println(count);
}
}
Boris told you how to make your program print the right answer, but the reason why it prints the right answer is, your program effectively is single threaded.
If you implemented Boris's suggestion, then your run() method probably looks like this:
public void run() {
synchronized (test1.class) {
for (int i = 0; i < 1000; i++) {
count++;
}
}
}
No two threads can ever be synchronized on the same object at the same time, and there's only one test1.class in your program. That's good because there's also only one count. You always want the number of lock objects and their lifetimes to match the number and lifetimes of the data that they are supposed to protect.
The problem is, you have synchronized the entire body of the run() method. That means, no two threads can run() at the same time. The synchronized block ensures that they all will have to execute in sequence—just as if you had simply called them one-by-one instead of running them in separate threads.
This would be better:
public void run() {
for (int i = 0; i < 1000; i++) {
synchronized (test1.class) {
count++;
}
}
}
If each thread releases the lock after each increment operation, then that gives other threads a chance to run concurrently.
On the other hand, all that locking and unlocking is expensive. The multi-threaded version almost certainly will take a lot longer to count to 10000 than a single threaded program would do. There's not much you can do about that. Using multiple CPUs to gain speed only works when there's big computations that each CPU can do independently of the others.
For your simple example, you can use AtomicInteger instead of static int and synchronized.
final AtomicInteger count = new AtomicInteger(0);
And inside Runnable only this one row:
count.IncrementAndGet();
Using syncronized blocks the whole class to be used by another threads if you have more complex codes with many of functions to use in a multithreaded code environment.
This code does'nt runs faster because of incrementing the same counter 1 by 1 is always a single operation which cannot run more than once at a moment.
So if you want to speed up running near 10x times faster, you should counting each thread it's own counter, than summing the results in the end. You can do this with ThreadPools using executor service and Future tasks wich can return a result for you.
This code I have is not executing tasks in parallel,
it only executes the code in this case once (whatever is in the for loop, but it should be 2) :
public class mqDirect {
public static void main(String args[]) throws Exception {
int parallelism = 2;
ExecutorService executorService =
Executors.newFixedThreadPool(parallelism);
Semaphore semaphore = new Semaphore(parallelism);
for (int i = 0; i < 1; i++) {
try {
semaphore.acquire();
// snip ... do stuff..
semaphore.release();
} catch (Throwable throwable) {
semaphore.release();
}
executorService.shutdownNow();
}
}
}
In Java the main way to make code work in parallel is to create a Thread with a new Runnable as a constructor parameter. You then need to start it.
There are many tutorials to help you get this to happen properly.
As your code stands you are merely creating an ExecutorService (and not using it), creating a Semaphore (which should be done in the thread but isn't), performing some process and then shutting down the Executor.
BTW: ShutDownNow is probably not what you want, you should just use ShutDown.
OK, So I found this good tutorial
http://programmingexamples.wikidot.com/threadpoolexecutor
And I have done something like
public class mqDirect {
int poolSize = 2;
int maxPoolSize = 2;
long keepAliveTime = 10;
ThreadPoolExecutor threadPool = null;
final ArrayBlockingQueue<Runnable> queue = new ArrayBlockingQueue<Runnable>(
5);
public mqDirect()
{
threadPool = new ThreadPoolExecutor(poolSize, maxPoolSize,
keepAliveTime, TimeUnit.SECONDS, queue);
}
public void runTask(Runnable task)
{
threadPool.execute(task);
System.out.println("Task count.." + queue.size());
}
public void shutDown()
{
threadPool.shutdown();
}
public static void main (String args[]) throws Exception
{
mqDirect mtpe = new mqDirect();
// start first one
mtpe.runTask(new Runnable()
{
public void run()
{
for (int i = 0; i < 2; i++)
{
try
{
System.out.println("First Task");
runMqTests();
Thread.sleep(1000);
} catch (InterruptedException ie)
{
}
}
}
});
// start second one
/*
* try{ Thread.sleep(500); }catch(InterruptedException
* ie){}
*/
mtpe.runTask(new Runnable()
{
public void run()
{
for (int i = 0; i < 2; i++)
{
try
{
System.out.println("Second Task");
runMqTests();
Thread.sleep(1000);
} catch (InterruptedException ie)
{
}
}
}
});
mtpe.shutDown();
// runMqTests();
}
And it works !
But the problem is , this duplicated code ... runMqtests() is the same task, is there a way to specify it to run in parallel without duplicating the code?
The example I based this off is assuming each task is different.
This code I have is not executing tasks in parallel, it only executes the code in this case once (whatever is in the for loop, but it should be 2) :
Just because you instantiate an ExecutorService instance doesn't mean that things magically run in parallel. You actually need to use that object aside from just shutting it down.
If you want the stuff in the loop to run in the threads in the service then you need to do something like:
int parallelism = 2;
ExecutorService executorService = Executors.newFixedThreadPool(parallelism);
for (int i = 0; i < parallelism; i++) {
executorService.submit(() -> {
// the code you want to be run by the threads in the exector-service
// ...
});
}
// once you have submitted all of the jobs, you can shut it down
executorService.shutdown();
// you might want to call executorService.awaitTermination(...) here
It is important to note that this will run your code in the service but there are no guarantees that it will be run "in parallel". This depends on your number of processors and the race conditions inherent with threads. For example, the first task might start up, run, and finish its code before the 2nd one starts. That's the nature of threaded programs which are by design asynchronous.
If, however, you have at least 2 cores, and the code that you submit to be run by the executor-service takes a long time to run then most likely they will be running at the same time at some point.
Lastly, as #OldCurmudgeon points out, you should call shutdown() on the service which allows current jobs already submitted to the service to run as opposed to shutdownNow() which cancels and queued jobs and also calls thread.interrupt() on any running jobs.
Hope this helps.
I'm pretty new using Threads and I just did a little program to understand how it works. As a example a did a prime numbers exercise, the program is running perfectly but what I have discovered is that if I don't use sleep(), the order of the numbers change everytime I press run (without changing the code). So why is happening that?
class Prime extends ThreadDemo implements Runnable
{
public void run()
{
for(int i=2;i<=20;i++)
{
if(prime(i))
{
System.out.printf ("Prime No.= %d \n",i);
}
}
}
}
class notPrime extends ThreadDemo implements Runnable
{
int number= 0;
public void run()
{
prime(number);
}
}
class ThreadDemo
{
public boolean prime(int start_value)
{
for(int i=2; i<start_value; i++)
{
if(start_value%i == 0)
{
System.err.printf ("No. Prime = %d \n", start_value);
return false;
}
}
return true;
}
public static void main(String args[])
{
Prime th1 = new Prime();
Thread childOne = new Thread(th1);
childOne.start();
notPrime th2 = new notPrime();
Thread childTwo = new Thread(th2);
childTwo.start();
}
}
This is the result after I pressed run:
This is the result after pressing again run:
The reason this is happening is that threads run in parallel. When you create a bunch of threads, these threads all start doing things at the same time, and it's a race to see which ones finish first. This isn't deterministic, and sometimes the threads will finish in a different order.
The reason sleep could change this is that sleep will give the threads you created first a head start.
There are two reasons why this happens :
Your prime method is not synchronized.
Even if you synchronize the prime method, you are passing different instances of ThreadDemo to your threads. Locks are obtained on objects. If two threads are passed two different objects of ThreadDemo, each thread will lock its own instance of ThreadDemo thus allowing the Threads to run in parallel.
There are a couple of changes you need to make to your code to make sure that the same ThreadDemo is used for both your threads.
class NotPrimeRunnable implements Runnable {
private ThreadDemo threadDemo;
int number = 0;
public NotPrimeRunnable(ThreadDemo threadDemo) {
this.threadDemo = threadDemo;
}
public void run() {
threadDemo.prime(number);
}
}
class PrimeRunnable implements Runnable {
private ThreadDemo threadDemo;
public PrimeRunnable(ThreadDemo threadDemo) {
this.threadDemo = threadDemo;
}
#Override
public void run() {
for (int i = 2; i <= 20; i++) {
if (threadDemo.prime(i)) {
System.out.printf("Prime No.= %d \n", i);
}
}
}
}
class ThreadDemo {
public synchronized boolean prime(int start_value) {
for (int i = 2; i < start_value; i++) {
if (start_value % i == 0) {
System.err.printf("No. Prime = %d \n", start_value);
return false;
}
}
return true;
}
public static void main(String args[]) {
ThreadDemo runnableTask = new ThreadDemo();
PrimeRunnable th1 = new PrimeRunnable(runnableTask);
Thread childOne = new Thread(th1);
childOne.start();
NotPrimeRunnable th2 = new NotPrimeRunnable(runnableTask);
Thread childTwo = new Thread(th2);
childTwo.start();
}
}
This will fix your problem.
Wikipedia defines thread as: "In computer science, a thread of execution is the smallest sequence of programmed instructions that can be managed independently by a scheduler".
Schedulers are software modules that run in the SO Kernel and manages the processes and threads execution. Schedulers use time-division multiplexing (as in multitasking) algorithms to ensure that all threads and process will be able to execute. One of the scheduler`s characteristics in SOs such as Windows, Mac OS and Linux is that the CPU switches between different software threads. As stated in Wikipedia: "This context switching generally happens frequently enough that the user perceives the threads or tasks as running at the same time."
Based on these considerations, we can explain your software behavior. Non-real time Operating System such as windows and MAC OSx, and many Linux distributions use scheduler algorithms that are non deterministic, so we cant predict the execution order of the threads, then it`s likely every time you execute you are going to get a different result regarding your text output order.
When you use sleep between the threads execution, it seems that the time amount chosen is enough to th1 execute completely before th2 starts. So the output is shown in the correct order.
I have hundreds of files to process. I do each file one at a time and it takes 30 minutes.
I'm thinking I can do this processing in 10 simultaneous threads, 10 files at a time, and I might be able to do it in 3 minutes instead of 30.
My question is, what is the "correct" way to manage my 10 threads? And when one is done, create a new one to a max number of 10.
This is what I have so far ... is this the "correct" way to do it?
public class ThreadTest1 {
public static int idCounter = 0;
public class MyThread extends Thread {
private int id;
public MyThread() {
this.id = idCounter++;
}
public void run() {
// this run method represents the long-running file processing
System.out.println("I'm thread '"+this.id+"' and I'm going to sleep for 5 seconds!");
try {
Thread.sleep(5000);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("I'm thread '"+this.id+"' and I'm done sleeping!");
}
}
public void go() {
int MAX_NUM_THREADS = 10;
List<MyThread> threads = new ArrayList<MyThread>();
// this for loop represents the 200 files that need to be processed
for (int i=0; i<200; i++) {
// if we've reached the max num of threads ...
while (threads.size() == MAX_NUM_THREADS) {
// loop through the threads until we find a dead one and remove it
for (MyThread t : threads) {
if (!t.isAlive()) {
threads.remove(t);
break;
}
}
}
// add new thread
MyThread t = new MyThread();
threads.add(t);
t.start();
}
}
public static void main(String[] args) {
new ThreadTest1().go();
}
}
You can use ExecutorService to manage you threads.
And you can add while loop to thread run method to execute file processing task repeatedly.
Also you can read about BlockingQueue usage. I think it will fit perfectly to allocate new files (tasks) between threads.
I would suggest using Camel's File component if you are open to it. The component will handle all the issues with concurrency to ensure that multiple threads don't try to process the same file. The biggest challenge with making your code multi-threaded is making sure the threads don't interact. Let a framework take care of this for you.
Example:
from("file://incoming?maxMessagesPerPoll=1&idempotent=true&moveFailed=failed&move=processed&readLock=none")
.threads(10).process()
I want to have two separate threads running two different instances of different classes and I want them to execute the run command at the same time.
I've made a practice class to demonstrate the problem I'm having.
One racer counts forwards, the other counts backwards.
public class testCount {
public static void main(String args[]) {
testCount countCompetition = new testCount();
countCompetition.run();
}
public void run() {
(new Thread(new racer1())).start();
(new Thread(new racer2())).start();
}
public class racer1 implements Runnable {
public void run() {
for(int x = 0; x < 100; x++) {
System.out.println(x);
}
}
}
public class racer2 implements Runnable {
public void run() {
for(int y = 100; y > 0; y--) {
System.out.println(y);
}
}
}
}
My results
1
2
... All the way to 100
100
100
99
... All the way back down
1
What I want
1
100
2
99
3
98
They don't need to be taking turns like that, but they do need to be working at the same time, instead of one after the other.
Any hints, advice or code snippets would be greatly appreciated.
I think all the answers so far are missing the point.
Your existing logic does enable your two threads to both execute concurrently, but this is not evident because your numbers only go up to 100, and the execution will usually stay with a specific thread for more than 1 instruction at a time, otherwise there would be a large amount of overhead in switching between the currently executing thread all the time. In your case, the JVM is deciding to execute your first thread long enough for it to print out 100 numbers before "context switching" to the 2nd thread. The JVM might choose to execute the threads differently, so the result you are seeing is not guaranteed to be the same every time.
If you increase your numbers even to 1000 you will (probably) see the two threads interleaving somewhat. You will still have large runs where one thread prints out a lot of numbers in a row because it is more efficient for the JVM to execute one thread for a while before switching, instead of context switching between every instruction.
Adding Thread.sleep(1) is not a good solution as you are adding an unneccessary delay. Sure, for 100 numbers this might not be noticable but for 10000 numbers you would have a delay of 10 seconds.
Is there any reason that you would require them to interleave to a higher degree than they already do? If there is then your simple model of running two threads concurrently is not sufficient. If not then just let the JVM decide the best order to run your threads in (which in the simple example you have given, means they probably won't interleave most of the time).
Just add Thread.sleep(1); in each racer class after System.out.println().
i.e. it will look like this:
public class racer1 implements Runnable {
public void run() {
for(int x = 0; x < 100; x++) {
System.out.println(x);
try {
Thread.sleep(1);
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
}
}
}
You need to write a basic wait and notify system. One task needs to notify the other that he has fished the work. Basic idea can be derived from below code. create 2 tasks, one to count forward and one to count backward
Runnable task = new Runnable() {
public void run() {
System.out.println("woohooTwo");
synchronized (t) {
while (true) {
System.out.println("---" + Thread.currentThread().getName() + "--" + t.i.getAndIncrement());
t.notifyAll();
try {
Thread.sleep(1000);
t.wait();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
};