This is the program
public class Thread2 implements Runnable {
private static int runTill = 10000;
private static int count = 0;
#Override
public void run() {
for(int i=0;i<runTill;i++) {
count++;
}
}
public static void main(String s[]) {
int iteration = 10;
for(int i = 0; i < iteration ;i++) {
Thread t = new Thread(new Thread2());
t.start();
}
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("Expected : "+(iteration * runTill));
System.out.println("Actual : "+count);
}
}
At the end I want count to be equal to (Expected : 100000). How can I achieve this?
A call to count++ is not atomic: it first has to load count, increment it and then store the new value in the variable. Without synchronization in place, threads will interleave during execution of this operation.
A simple way to get what you want is to use an AtomicInteger:
private static AtomicInteger count = new AtomicInteger();
#Override
public void run() {
for(int i=0;i<runTill;i++) {
count.incrementAndGet();
}
}
use "compare and set" instead of "increment and get"
private static AtomicInteger count = new AtomicInteger();
#Override
public void run() {
for(int i=0;i<runTill;i++) {
//note: another thread might reach this point at the same time when i is 9,999
// (especially if you have other codes running prior to the increment within the for loop)
// then count will be added 2x if you use incrementAndGet
boolean isSuccessful = count.compareAndSet(i, i+1);
if(!isSuccessful)
System.out.println("number is not increased (another thread already updated i)");
}
}
As the comments suggest, besides the need for synchronizing access (to count, became an AtomicInteger here), threads should be waited to complete using Thread.join(), instead of "guessing" their runtime:
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.atomic.AtomicInteger;
public class Thread2 implements Runnable {
private static int runTill = 10000;
private static AtomicInteger count = new AtomicInteger();
#Override
public void run() {
for (int i = 0; i < runTill; i++) {
count.incrementAndGet();
}
}
public static void main(String s[]) {
int iteration = 10;
List<Thread> threads = new ArrayList<Thread>();
for (int i = 0; i < iteration; i++) {
Thread t = new Thread(new Thread2());
threads.add(t);
t.start();
}
try {
for (Thread t : threads)
t.join();
} catch (InterruptedException ie) {
ie.printStackTrace();
}
System.out.println("Expected : " + (iteration * runTill));
System.out.println("Actual : " + count);
}
}
Hi I am trying to print even and odd using two threads namedly EvenThread and OddThread, some times I am getting correct result and some times not, could any one please help me.
package com.java8;
public class EvenOddExample {
public static synchronized void print(int i,String name){
System.out.println(i+"--->"+name);
}
public static void main(String[] args) throws InterruptedException {
EvenThread e= new EvenThread();
e.start();
OddThread o=new OddThread();
o.start();
}
public static class EvenThread extends Thread{
public void run() {
for(int i=0;i<10;i++){
if(i%2==0){
print(i,"Even");
}else{
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
}
public static class OddThread extends Thread{
#Override
public void run() {
for(int i=1;i<10;i++){
if(i%2!=0){
print(i,"Odd");
}else{
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
}
}
You need some signaling between the two threads. Putting synchronized on the print method simply guarantees, that only one thread can enter the method at a time. To put your threads into order Object.wait() and Object.notify{All}() methods can be used.
Actually this is some kind of the Sender-Receiver Synchronization Problem. Based on the example of the problem described here (Please read this page in order to understand how this synchronization works) I adapted your code. Additionally I used ExecutorService and Callable instead of extending Thread, which is bad-practice:
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
public class EvenOddExample {
private static boolean evensTurn = true;
private static Object monitor = new Object();
public static void print(int i, String name) {
System.out.println(i + "--->" + name);
}
public static void main(String[] args) throws InterruptedException {
final ExecutorService executorService = Executors.newFixedThreadPool(2);
executorService.submit(new EvenCallable());
executorService.submit(new OddCallable());
executorService.shutdown();
}
public static class EvenCallable implements Callable<Void> {
#Override
public Void call() throws InterruptedException {
for (int i = 0; i < 10; i++) {
if (i % 2 == 0) {
synchronized (monitor) {
while (!evensTurn) { // not your turn?
monitor.wait(); // wait for monitor in a loop to handle spurious wakeups
}
print(i, "Even");
evensTurn = false; // next odd needs to run
monitor.notifyAll(); // wakeup the odd thread
}
} else {
Thread.sleep(1000);
}
}
return null;
}
}
public static class OddCallable implements Callable<Void> {
#Override
public Void call() throws InterruptedException {
for (int i = 1; i < 10; i++) {
if (i % 2 != 0) {
synchronized (monitor) {
while (evensTurn) {
monitor.wait();
}
print(i, "Odd");
evensTurn = true;
monitor.notifyAll();
}
} else {
Thread.sleep(1000);
}
}
return null;
}
}
}
synchronized is used to lock the access of another thread, when the locked object is free, it does not guarantee which is next called thread. You can use semaphore to make inter-thread communication:
private static Semaphore[] semaphores = {new Semaphore(0), new Semaphore(1)};
static void print(int i, String name) {
try {
semaphores[(i + 1) % 2].acquire();
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
System.out.println(i + "--->" + name);
semaphores[i % 2].release();
}
public class EvenOddPrinter {
static boolean flag = true;
public static void main(String[] args) {
class Odd implements Runnable {
#Override
public void run() {
for (int i = 1; i <= 10;) {
if (EvenOddPrinter.flag) {
System.out.println(i + "--->odd");
i += 2;
EvenOddPrinter.flag = !EvenOddPrinter.flag;
}
}
}
}
class Even implements Runnable {
#Override
public void run() {
for (int i = 2; i <= 10;) {
if (!EvenOddPrinter.flag) {
System.out.println(i + "---->even");
i += 2;
EvenOddPrinter.flag = !EvenOddPrinter.flag;
}
}
}
}
Runnable odd = new Even();
Runnable even = new Odd();
Thread t1 = new Thread(odd, "Odd");
Thread t2 = new Thread(even, "Even");
t1.start();
t2.start();
}
}
There are two thread, A Thread let the list add ten times, B Thread will quit when the list.size() > 5.I think the result should be "B would quit after A add 5, maybe 6,7,7,8,9",but the result is B never quit.
and the other question is when i add a line there, it would be ok! why?
package com.lock;
import java.util.*;
import java.util.concurrent.CopyOnWriteArrayList;
/**
* Create by #author Henry on 2018/5/4
*/
public class NoWait {
static volatile ArrayList list = new ArrayList();
public static void main(String[] args) throws InterruptedException {
ThreadA a = new ThreadA(list);
a.setName("A");
a.start();
ThreadB b = new ThreadB(list);
b.setName("B");
b.start();
Thread.sleep(10000);
System.out.println("A is " + a.isAlive());
System.out.println("B is " + b.isAlive());
}
}
class ThreadA extends Thread{
private ArrayList list;
public ThreadA(ArrayList list){
this.list = list;
}
#Override
public void run(){
try{
for (int i = 0; i < 10; i++) {
list.add(i);
System.out.println("=====================================add "+(i+1));
Thread.sleep(100);
}
} catch (Exception e) {
e.printStackTrace();
}
System.out.println("A finished");
}
}
class ThreadB extends Thread{
String s = "";
private ArrayList list;
public ThreadB(ArrayList list){
this.list = list;
}
#Override
public void run(){
try{
while(true){
//s += list.size();
//System.out.println(list.size());
if(list.size() > 5){
System.out.println(">5 now, B gonne quit");
throw new InterruptedException();
}
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
the result is B Thread never stop! this is my first question.
the second is , when I add this line
//s += list.size();
//System.out.println(list.size());
and it will be ok, Why?!! how they works!
Thread B is not guaranteed to see any changes from thread A because there is no synchronization between them.
Volatile won't do anything here as it only ensures the reference to the list is visible to both threads but not updates to its contents.
The println has internal synchronization which causes the changes between the threads to be visible.
I have a following code as below:
class Example {
private volatile int testValue = 0;
public int getTestValue() {
return testValue;
}
public void setTestValue(int testValue) {
this.testValue = testValue;
}
public void increment() {
this.testValue += 1;
}
}
class PrintThread extends Thread {
private Example example;
private int x = 0;
public PrintThread(Example example) {
this.example = example;
x = example.getTestValue();
}
public void run() {
while(true) {
if(x != example.getTestValue()) { // block 1
System.out.println("printThread: " + example.getTestValue());
x = example.getTestValue();
}
}
}
}
class IncrementorThread extends Thread {
private Example example;
public IncrementorThread(Example example) {
this.example = example;
}
public void run() {
while(true) {
example.increment();
System.out.println("incrementorThread: " + example.getTestValue());
try {
Thread.sleep(800);
} catch(Exception ex) {
}
}
}
}
public class VolatileExample {
public static void main(String args[]) {
Example ex = new Example();
new IncrementorThread(ex).start();
new PrintThread(ex).start();
}
}
When I remove volatile keyword in Example class then I never see the output of PrintThread. In PrintThread when I print out the testValue of example, value of example object still updated but the code in 'block 1' never be executed. Both thread still access the same object, can anyone explain me more detail about this? About the volatile keyword affected in this case
You should use atomic integers insteed of volatile fields. To get the idea why that is important try running code below. Here you have 3 types of variables, normal int, volatile int and AtomicInteger. Only AtomicInteger assure the thread safety of value. After running this simple code, you will see why.
public class Test {
private int threadCount = 10;
private int nonVolatileCount = 0;
private volatile int volatileCount = 0;
private AtomicInteger atomicCount = new AtomicInteger(0);
private CountDownLatch startLatch = new CountDownLatch(threadCount);
private CountDownLatch endLatch = new CountDownLatch(threadCount);
private class Task implements Runnable {
public void run() {
startLatch.countDown();
try {
startLatch.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int i = 0; i < 1000000; i++) {
nonVolatileCount++;
volatileCount++;
atomicCount.incrementAndGet();
}
endLatch.countDown();
};
}
public static void main(String[] args) throws InterruptedException {
new Test().go();
}
public void go() throws InterruptedException {
for (int i = 0; i < threadCount; i++) {
new Thread(new Task()).start();
}
endLatch.await();
System.out.println("non volatile counter: " + nonVolatileCount);
System.out.println(" volatile counter: " + volatileCount);
System.out.println(" atomic counter: " + atomicCount.get());
}
}
How can i order threads in the order they were instantiated.e.g. how can i make the below program print the numbers 1...10 in order.
public class ThreadOrdering {
public static void main(String[] args) {
class MyRunnable implements Runnable{
private final int threadnumber;
MyRunnable(int threadnumber){
this.threadnumber = threadnumber;
}
public void run() {
System.out.println(threadnumber);
}
}
for(int i=1; i<=10; i++){
new Thread(new MyRunnable(i)).start();
}
}
}
Sounds like you want ExecutorService.invokeAll, which will return results from worker threads in a fixed order, even though they may be scheduled in arbitrary order:
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
public class ThreadOrdering {
static int NUM_THREADS = 10;
public static void main(String[] args) {
ExecutorService exec = Executors.newFixedThreadPool(NUM_THREADS);
class MyCallable implements Callable<Integer> {
private final int threadnumber;
MyCallable(int threadnumber){
this.threadnumber = threadnumber;
}
public Integer call() {
System.out.println("Running thread #" + threadnumber);
return threadnumber;
}
}
List<Callable<Integer>> callables =
new ArrayList<Callable<Integer>>();
for(int i=1; i<=NUM_THREADS; i++) {
callables.add(new MyCallable(i));
}
try {
List<Future<Integer>> results =
exec.invokeAll(callables);
for(Future<Integer> result: results) {
System.out.println("Got result of thread #" + result.get());
}
} catch (InterruptedException ex) {
ex.printStackTrace();
} catch (ExecutionException ex) {
ex.printStackTrace();
} finally {
exec.shutdownNow();
}
}
}
"I actually have some parts that i want to execute in parallel, and then once the results are generated, I want to merge the results in certain order." Thanks, this clarifies what you're asking.
You can run them all at once, but the important thing is to get their results in order when the threads finish their computation. Either Thread#join() them in the order in which you want to get their results, or just Thread#join() them all and then iterate through them to get their results.
// Joins the threads back to the main thread in the order we want their results.
public class ThreadOrdering {
private class MyWorker extends Thread {
final int input;
int result;
MyWorker(final int input) {
this.input = input;
}
#Override
public void run() {
this.result = input; // Or some other computation.
}
int getResult() { return result; }
}
public static void main(String[] args) throws InterruptedException {
MyWorker[] workers = new MyWorker[10];
for(int i=1; i<=10; i++) {
workers[i] = new MyWorker(i);
workers[i].start();
}
// Assume it may take a while to do the real computation in the threads.
for (MyWorker worker : workers) {
// This can throw InterruptedException, but we're just passing that.
worker.join();
System.out.println(worker.getResult());
}
}
}
Simply put, the scheduling of threads is indeterminate.
http://www.janeg.ca/scjp/threads/scheduling.html Dead domain - do not click
WaybackMachine version of the above page
The longer answer is that if you want to do this, you'll need to manually wait for each thread to complete its work before you allow another to run. Note that in this fashion, all the threads will still interleave but they won't accomplish any work until you give the go-ahead. Have a look at the synchronize reserved word.
You can chain them – that is, have the first one start the second, the second start the third, etc. They won't really be running at the same time except for a bit of overlap when each one is started.
public class ThreadOrdering
{
public static void main(String[] args)
{
MyRunnable[] threads = new MyRunnable[10];//index 0 represents thread 1;
for(int i=1; i<=10; i++)
threads[i] = new MyRunnable(i, threads);
new Thread(threads[0].start);
}
}
public class MyRunnable extends Runnable
{
int threadNumber;
MyRunnable[] threads;
public MyRunnable(int threadNumber, MyRunnable[] threads)
{
this.threadnumber = threadnumber;
this.threads = threads;
}
public void run()
{
System.out.println(threadnumber);
if(threadnumber!=10)
new Thread(threadnumber).start();
}
}
Here's a way to do it without having a master thread that waits for each result. Instead, have the worker threads share an object which orders the results.
import java.util.*;
public class OrderThreads {
public static void main(String... args) {
Results results = new Results();
new Thread(new Task(0, "red", results)).start();
new Thread(new Task(1, "orange", results)).start();
new Thread(new Task(2, "yellow", results)).start();
new Thread(new Task(3, "green", results)).start();
new Thread(new Task(4, "blue", results)).start();
new Thread(new Task(5, "indigo", results)).start();
new Thread(new Task(6, "violet", results)).start();
}
}
class Results {
private List<String> results = new ArrayList<String>();
private int i = 0;
public synchronized void submit(int order, String result) {
while (results.size() <= order) results.add(null);
results.set(order, result);
while ((i < results.size()) && (results.get(i) != null)) {
System.out.println("result delivered: " + i + " " + results.get(i));
++i;
}
}
}
class Task implements Runnable {
private final int order;
private final String result;
private final Results results;
public Task(int order, String result, Results results) {
this.order = order;
this.result = result;
this.results = results;
}
public void run() {
try {
Thread.sleep(Math.abs(result.hashCode() % 1000)); // simulate a long-running computation
}
catch (InterruptedException e) {} // you'd want to think about what to do if interrupted
System.out.println("task finished: " + order + " " + result);
results.submit(order, result);
}
}
If you need that fine-grained control, you should not use threads but instead look into using a suitable Executor with Callables or Runnables. See the Executors class for a wide selection.
A simple solution would be to use an array A of locks (one lock per thread). When thread i begins its executions, it acquires its associated lock A[i]. When it's ready to merge its results, it releases its lock A[i] and waits for locks A[0], A[1], ..., A[i - 1] to be released; then it merges the results.
(In this context, thread i means the i-th launched thread)
I don't know what classes to use in Java, but it must be easy to implement. You can begin reading this.
If you have more questions, feel free to ask.
public static void main(String[] args) throws InterruptedException{
MyRunnable r = new MyRunnable();
Thread t1 = new Thread(r,"A");
Thread t2 = new Thread(r,"B");
Thread t3 = new Thread(r,"C");
t1.start();
Thread.sleep(1000);
t2.start();
Thread.sleep(1000);
t3.start();
}
Control of thread execution order may be implemented quite easily with the semaphores. The code attached is based on the ideas presented in Schildt's book on Java (The complete reference....).
// Based on the ideas presented in:
// Schildt H.: Java.The.Complete.Reference.9th.Edition.
import java.util.concurrent.Semaphore;
class Manager {
int n;
// Initially red on semaphores 2&3; green semaphore 1.
static Semaphore SemFirst = new Semaphore(1);
static Semaphore SemSecond = new Semaphore(0);
static Semaphore SemThird = new Semaphore(0);
void firstAction () {
try {
SemFirst.acquire();
} catch(InterruptedException e) {
System.out.println("Exception InterruptedException catched");
}
System.out.println("First: " );
System.out.println("-----> 111");
SemSecond.release();
}
void secondAction() {
try{
SemSecond.acquire();
} catch(InterruptedException e) {
System.out.println("Exception InterruptedException catched");
}
System.out.println("Second: ");
System.out.println("-----> 222");
SemThird.release();
}
void thirdAction() {
try{
SemThird.acquire();
} catch(InterruptedException e) {
System.out.println("Exception InterruptedException catched");
}
System.out.println("Third: ");
System.out.println("-----> 333");
SemFirst.release();
}
}
class Thread1 implements Runnable {
Manager q;
Thread1(Manager q) {
this.q = q;
new Thread(this, "Thread1").start();
}
public void run() {
q.firstAction();
}
}
class Thread2 implements Runnable {
Manager q;
Thread2(Manager q) {
this.q = q;
new Thread(this, "Thread2").start();
}
public void run() {
q.secondAction();
}
}
class Thread3 implements Runnable {
Manager q;
Thread3(Manager q) {
this.q = q;
new Thread(this, "Thread3").start();
}
public void run() {
q.thirdAction();
}
}
class ThreadOrder {
public static void main(String args[]) {
Manager q = new Manager();
new Thread3(q);
new Thread2(q);
new Thread1(q);
}
}
This can be done without using synchronized keyword and with the help of volatile keyword. Following is the code.
package threadOrderingVolatile;
public class Solution {
static volatile int counter = 0;
static int print = 1;
static char c = 'A';
public static void main(String[] args) {
// TODO Auto-generated method stub
Thread[] ths = new Thread[4];
for (int i = 0; i < ths.length; i++) {
ths[i] = new Thread(new MyRunnable(i, ths.length));
ths[i].start();
}
}
static class MyRunnable implements Runnable {
final int thID;
final int total;
public MyRunnable(int id, int total) {
thID = id;
this.total = total;
}
#Override
public void run() {
while(true) {
if (thID == (counter%total)) {
System.out.println("thread " + thID + " prints " + c);
if(c=='Z'){
c='A';
}else{
c=(char)((int)c+1);
}
System.out.println("thread " + thID + " prints " + print++);
counter++;
} else {
try {
Thread.sleep(30);
} catch (InterruptedException e) {
// log it
}
}
}
}
}
}
Following is the github link which has a readme, that gives detailed explanation about how it happens.
https://github.com/sankar4git/volatile_thread_ordering