public class LockExample {
public static void main(String[] args) {
SharedResource sharedResource = new SharedResource();
Thread A = new Thread(()->{for(int i = 0; i < 10; i++)sharedResource.print5();},"A");
Thread B = new Thread(()->{for(int i = 0; i < 10; i++)sharedResource.print10();},"B");
Thread C = new Thread(()->{for(int i = 0; i < 10; i++)sharedResource.print15();},"C");
A.start();
B.start();
C.start();
try {
A.join();
B.join();
C.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("totall calls: "+sharedResource.getTotalCalls());
}
}
class SharedResource{
private int number = 1;
private int totalCalls = 0;
private Lock lock = new ReentrantLock();
private Condition condition1 = lock.newCondition();
private Condition condition2 = lock.newCondition();
private Condition condition3 = lock.newCondition();
public void print5(){
lock.lock();
try {
System.out.println("print5 running");
while(number != 2)condition1.await();
for(int i =0; i<5; i++){
System.out.println(Thread.currentThread().getName()+"\t"+i);
}
totalCalls++;
number = 3;
condition3.signal();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
public void print10(){
lock.lock();
try {
System.out.println("print10 running");
while(number != 1)condition2.await();
for(int i =0; i<10; i++){
System.out.println(Thread.currentThread().getName()+"\t"+i);
}
totalCalls++;
number = 2;
condition1.signal();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
public void print15(){
lock.lock();
try {
System.out.println("print15 running");
while(number != 3)condition3.await();
for(int i =0; i<15; i++){
System.out.println(Thread.currentThread().getName()+"\t"+i);
}
totalCalls++;
number = 1;
condition2.signal();
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
public int getTotalCalls() {
return totalCalls;
}
}
Hello guys, consider the code above, I thought the lock must lock the critical section that been surrounded by lock and unlock, but why the output look like this?
print5 running
print10 running
B 0
B 1
B 2
B 3
B 4
B 5
B 6
B 7
B 8
B 9
print10 running
print15 running
A 0
A 1
Why it go back to println("print10 running") again at output line 13?
I think it signal the next thread which is print15() then print15() get the lock and lock everything. Hence print10 can not go in to lock section anymore. So why it print "print10 running" which is inside the lock section again?
My first question, Thank for your help!
I'm trying to print odd and even numbers 1~100 alternatively using two threads.
Expected results：
pool-1-thread-1=> 1
pool-1-thread-2=> 2
pool-1-thread-1=> 3
pool-1-thread-2=> 4
......
pool-1-thread-1=> 99
pool-1-thread-2=> 100
I think i can use FairSync, but it can only guarantee that most of the print is correct. like this：
pool-1-thread-1=> 55
pool-1-thread-2=> 56
pool-1-thread-1=> 57
pool-1-thread-2=> 58
pool-1-thread-2=> 59 //※error print※
pool-1-thread-1=> 60
pool-1-thread-2=> 61
pool-1-thread-1=> 62
I don't know why is the order lost in very few cases?
You can criticize my code and my English.
Here is my code:
private static final int COUNT = 100;
private static final int THREAD_COUNT = 2;
private static int curr = 1;
static ReentrantLock lock = new ReentrantLock(true);
static ExecutorService executorService = Executors.newCachedThreadPool();
public static void main(String[] args) {
Runnable task = () -> {
for (; ; ) {
try {
lock.lock();
if (curr <= COUNT) {
System.out.println(Thread.currentThread().getName() + "=> " + curr++);
} else {
System.exit(0);
}
} catch (Exception e) {
e.printStackTrace();
} finally {
lock.unlock();
}
}
};
for (int i = 0; i < THREAD_COUNT; i++) {
executorService.execute(task);
}
}
No dear your implementation is not correct. Which thread get's the opportunity to RUN is decided by the OS. Thread 1 & 2 will execute one after another cannot be guaranteed.
You can fix your code by checking the previous value of the variable curr and if the value is not what this thread expects don't increment and print.
for eg :
if(curr.threadName.equals("Thread 2") && (curr%2 !=0))
{
// Print
// Increment
}
You cant use single lock to achieve this. Even ReentrantLock gives fairness but it cant control thread schedule.
We can achieve throw inter thread communication like Semaphore. Semaphore controls the thread execution.
We create two threads, an odd thread, and an even thread. The odd thread would print the odd numbers starting from 1, and the even thread will print the even numbers starting from 2.
Create two semaphores, semOdd and semEven which will have 1 and 0 permits to start with. This will ensure that odd number gets printed first.
class SharedPrinter {
private Semaphore semEven = new Semaphore(0);
private Semaphore semOdd = new Semaphore(1);
void printEvenNum(int num) {
try {
semEven.acquire();
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
System.out.println(Thread.currentThread().getName() + num);
semOdd.release();
}
void printOddNum(int num) {
try {
semOdd.acquire();
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
}
System.out.println(Thread.currentThread().getName() + num);
semEven.release();
}
}
class Even implements Runnable {
private SharedPrinter sp;
private int max;
// standard constructor
#Override
public void run() {
for (int i = 2; i <= max; i = i + 2) {
sp.printEvenNum(i);
}
}
}
class Odd implements Runnable {
private SharedPrinter sp;
private int max;
// standard constructors
#Override
public void run() {
for (int i = 1; i <= max; i = i + 2) {
sp.printOddNum(i);
}
}
}
public static void main(String[] args) {
SharedPrinter sp = new SharedPrinter();
Thread odd = new Thread(new Odd(sp, 10),"Odd");
Thread even = new Thread(new Even(sp, 10),"Even");
odd.start();
even.start();
}
Refer : here
I am trying a write a simple program with wait and notify in which I will create 3 threads.
The first thread should print 1, 2, 3.
The second thread should print 4, 5, 6.
The third thread should print 7, 8, 9.
After that, the first thread should print 10, 11, 12 and so on.
Below is a sample code for the same exercise, but I am unable to print the desired output.
public class MyThread2 extends Thread {
public final static Object obj = new Object();
int threadNo;
static volatile int threadNoToRun;
static volatile int counter = 1;
public MyThread2(int threadNo){
this.threadNo= threadNo;
}
#Override
public void run() {
synchronized (obj) {
try {
if(threadNoToRun != threadNo)
obj.wait();
else{
for(int i = 1 ; i < 4 ; i++){
if(threadNoToRun == threadNo){
System.out.println(threadNo + " counter value is "+counter);
counter++;
System.out.println(threadNo + " counter value is "+counter);
counter++;
System.out.println(threadNo + " counter value is "+counter);
counter++;
if(threadNoToRun == 1){
threadNoToRun = 2;
}
else if(threadNoToRun == 2){
threadNoToRun = 3;
}
else if(threadNoToRun == 3){
threadNoToRun = 1;
}
}
}
obj.notifyAll();
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
public static void main (String args[]) {
/*
* Creating as many threads as needed.
*/
MyThread2 th1 = new MyThread2(1);
MyThread2 th2 = new MyThread2(2);
MyThread2 th3 = new MyThread2(3);
MyThread2.threadNoToRun = 1;
th1.start();
th2.start();
th3.start();
}
}
The output looks like:
1 counter value is 1
1 counter value is 2
1 counter value is 3
2 counter value is 4
2 counter value is 5
2 counter value is 6
Here, it was just a few changes.
Nonetheless, I have to point out that this kind of concurrency does NOT increase computation speed. Only one thread is alive at all times.
public class MyThread2 extends Thread {
public final static Object obj = new Object();
int threadNo;
static volatile int threadNoToRun;
static volatile int counter = 1;
public MyThread2(int threadNo) {
this.threadNo = threadNo;
}
#Override
public void run() {
synchronized (obj) {
try {
while (counter < 100) {
if (threadNoToRun != threadNo)
obj.wait();
else {
System.out.println(threadNo + " counter value is " + counter);
counter++;
System.out.println(threadNo + " counter value is " + counter);
counter++;
System.out.println(threadNo + " counter value is " + counter);
counter++;
if (threadNoToRun == 1) {
threadNoToRun = 2;
} else if (threadNoToRun == 2) {
threadNoToRun = 3;
} else if (threadNoToRun == 3) {
threadNoToRun = 1;
}
obj.notifyAll();
}
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
public static void main(String args[]) {
/*
* Creating as many threads as needed.
*/
MyThread2 th1 = new MyThread2(1);
MyThread2 th2 = new MyThread2(2);
MyThread2 th3 = new MyThread2(3);
MyThread2.threadNoToRun = 1;
th1.start();
th2.start();
th3.start();
}
}
I tried to print odd number in one thread and even number in another. I tried creating two thread and printing it in run method.
public class OddEven
{
private final int MAX = 10;
private static int counter = 0;
private volatile boolean isOdd = true;
public synchronized void printEven(int counter)
{
try {
if (!isOdd) {
System.out.println(Thread.currentThread().getName() + " " + counter);
counter++;
isOdd = true;
}
wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
public synchronized void printOdd(int counter)
{
if (isOdd) {
counter++;
System.out.println(Thread.currentThread().getName() + " " + counter);
isOdd = false;
}
notifyAll();
}
public static void main(String[] args) {
OddEven oddEven = new OddEven();
Thread th1 = new Thread() {
public void run() {
while (OddEven.counter < oddEven.MAX) {
oddEven.printEven(OddEven.counter);
}
}
};
th1.setName("even -");
th1.start();
Thread th2 = new Thread() {
public void run() {
while (OddEven.counter < oddEven.MAX) {
oddEven.printOdd(OddEven.counter);
}
}
};
th2.setName("odd -");
th2.start();
}
}
But it is printing it like below infinitely.
even - 0
odd - 1
even - 0
odd - 1
even - 0
odd - 1
To read: Is Java "pass-by-reference" or "pass-by-value"?
You pass in a primitive. counter++; makes sense only within the method and has no impact on the outer world. count refers to the method param, not to the field this.count.
There is no proper synchronisation placed upon the condition OddEven.counter < oddEven.MAX, so different things may happen.
My advice would be to remove isOdd and do a check on the spot. For instance,
public synchronized void printEven() {
if (counter % 2 != 0) {
System.out.println(Thread.currentThread().getName() + " " + ++counter);
}
}
The line oddEven.printEven(OddEven.counter) passes an integer by value to the printEven method which does not change the value of OddEven.counter when it does counter++ as also pointed in other answers here.
To get the desired output, one option is to remove the passed parameter to both printEven and printOdd methods. And there are many other ways to achieve what you are trying to do here.
And there is also a mistake in the printEven method. counter++; needs to be before the print statement.
This will give you the desired output.
I'm new to muti-threading and I got a question to print 1 to 100 using 10 threads in Java with below constrain.
Thread t1 should print:
1, 11, 21, 31, ... 91
t2 should print:
2, 12, 22, 32, ... 92
likewise
t10 should print:
10, 20, 30, ... 100
The final output should be
1 2 3 .. 100
I have tried it, but it is throwing the following exception in all 10 threads:
java.lang.IllegalMonitorStateException
at java.lang.Object.wait(Native Method)
at java.lang.Object.wait(Object.java:485)
at thread.run(MyThread.java:58)
at java.lang.Thread.run(Unknown Source)
Please let me know how I can solve this problem.
public class MyThread {
/**
* #param args
*/
public static void main(String[] args) {
thread.setSequence();
for(int i = 1; i <= 10; i++) {
Thread t = new Thread(new thread(i));
t.setName(i + "");
t.start();
}
}
}
class thread implements Runnable {
private static HashMap< String, String> sequence = new HashMap<String, String>();
public static final Object lock = new Object();
public static String turn = "1";
private int startValue = 0;
private AtomicInteger counter = new AtomicInteger(1);
public thread(int startValue){
this.startValue = startValue;
}
#Override
public void run() {
while (!counter.equals(10)){
synchronized (lock) {
if(Thread.currentThread().getName().equals(turn)){
System.out.print(startValue + " ");
startValue += 10;
counter.incrementAndGet();
turn = getNextTurn(turn);
try {
this.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
else{
try {
this.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
this.notifyAll();
}
}
}
public static void setSequence(){
for (int i = 1; i <= 10; i++)
if (i == 10)
sequence.put(i + "", 1 + "");
else
sequence.put(i + "", (i + 1) + "");
}
public static String getNextTurn(String currentTurn){
return sequence.get(currentTurn);
}
}
The simplest way would be to have a volatile variable from which each thread reads in and update according to its turn, otherwise it just waits until his turn. When counter is equals to 100 you stop all threads to run by breaking the outer loop.
class MyRunnable implements Runnable {
private static final int LIMIT = 20;
private static volatile int counter = 0;
private int id;
public MyRunnable(int id) {
this.id = id;
}
#Override
public void run() {
outer:
while(counter < LIMIT) {
while (counter % NB_THREADS != id) {
if(counter == LIMIT) break outer;
}
System.out.println("Thread "+Thread.currentThread().getName()+ " printed " + counter);
counter += 1;
}
}
}
Given a LIMIT of 20 and 10 threads, it outputs:
Thread 0 printed 0
Thread 1 printed 1
Thread 2 printed 2
Thread 3 printed 3
Thread 4 printed 4
Thread 5 printed 5
Thread 6 printed 6
Thread 7 printed 7
Thread 8 printed 8
Thread 9 printed 9
Thread 0 printed 10
Thread 1 printed 11
Thread 2 printed 12
Thread 3 printed 13
Thread 4 printed 14
Thread 5 printed 15
Thread 6 printed 16
Thread 7 printed 17
Thread 8 printed 18
Thread 9 printed 19
Of course, this is a very bad usage of multithreading because each thread waits its turn to print and increment the counter.
Multithreading works well when threads can work independently of another for relatively long time's window, and then may occasionally meet up to compare or combine their results if needed.
For example in the fork-join model, each thread does its task independently then their results are merged to produce the final outcome, such as in a merge sort for example. But this assume that the task can be easily parallelizable into independant subtasks, which is not the case here because your final output should be consecutive numbers.
So here a simple loop would be largely more efficient, but I can understand it's for learning purposes.
Here is a solution for the problem.The current thread acquire the lock and we decide if the thread is eligible to execute (printing the number here). If so perform the operation and notify all threads that they can try now. Else wait till its notified by other threads.
public class MyThread extends Thread{
//define the Total No.Of Threads needed
public static final int TOTAL_THREADS = 10;
public final static Object obj = new Object();
int threadNo;
static volatile int counter = 1;
public MyThread(int threadNo){
this.threadNo= threadNo;
}
#Override
public void run(){
//in a synchronized block to acquire lock
synchronized (obj) {
while(counter<=100){
/*
* counter==threadNo => To print the initial numbers till TOTAL_THREADS
* counter%TOTAL_THREADS == threadNo => e.g 11%10 = 1 -> 1 will print this, 12%10 = 2 ..
* (counter%TOTAL_THREADS == 0) && (TOTAL_THREADS == threadNo) => 10%10 will be 0,
* and this must be printed by 10 th thread only, ie the highest thread.
*/
if(counter == threadNo || (counter%TOTAL_THREADS == threadNo) ||
((counter%TOTAL_THREADS == 0) && (TOTAL_THREADS == threadNo))){
//Display the output as desired
System.out.println(this.threadNo+" printing"+" "+counter++);
//notify
obj.notifyAll();
}else{
//current thread not eligible for printing the current counter value, so wait till its notified
try {
obj.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
public static void main (String args[]) {
/*
* Creating as many threads as needed.
*/
for(int i = 1; i<=TOTAL_THREADS;i++){
MyThread th = new MyThread(i);
th.start();
}
}
}
The output will be
1 printing 1,
2 printing 2,
3 printing 3,
4 printing 4,
5 printing 5,
6 printing 6,
7 printing 7,
8 printing 8,
9 printing 9,
10 printing 10,
1 printing 11,
2 printing 12,
3 printing 13,
4 printing 14,
...
7 printing 97,
8 printing 98,
9 printing 99,
10 printing 100
Hope this helps =) Took me an hour to do it.
package com.xxxx.simpleapp;
import java.util.ArrayList;
import java.util.List;
public class TenThreads {
public int currentTaskValue = 1;
public static void main(String[] args) {
TenThreads monitor = new TenThreads();
List<ModThread> list = new ArrayList();
for (int i = 0; i < 10; i++) {
ModThread modThread = new ModThread(i, monitor);
list.add(modThread);
}
for (ModThread a : list) {
a.start();
}
}
}
class ModThread extends Thread {
private int modValue;
private TenThreads monitor;
public ModThread(int modValue, TenThreads monitor) {
this.modValue = modValue;
this.monitor = monitor;
}
#Override
public void run() {
synchronized (monitor) {
try {
while (true) {
while (monitor.currentTaskValue % 10 != modValue) {
monitor.wait();
}
if (monitor.currentTaskValue == 101) {
break;
}
System.out.println(Thread.currentThread().getName() + " : "
+ monitor.currentTaskValue + " ,");
monitor.currentTaskValue = monitor.currentTaskValue + 1;
monitor.notifyAll();
}
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
output
Thread-1 : 1 ,
Thread-2 : 2 ,
Thread-3 : 3 ,
Thread-4 : 4 ,
Thread-5 : 5 ,
Thread-6 : 6 ,
Thread-7 : 7 ,
Thread-8 : 8 ,
Thread-9 : 9 ,
......
.....
...
Thread-4 : 94 ,
Thread-5 : 95 ,
Thread-6 : 96 ,
Thread-7 : 97 ,
Thread-8 : 98 ,
Thread-9 : 99 ,
Thread-0 : 100 ,
Documentation are intentionally left out for you to figure it out, there are minor bugs too!
Error is thrown due to calling of wait not on proper object. wait() should be called on object on which lock is acquired, the one implied by synchronized keyword.
Well I do not have the code...but the perspective seems to be
that there are 100 tasks to be executed each of incrementing
a count by 1.
So there could be a ThreadPool of say 10 threads and these
threads are incrementing the shared count value...
Only point to consider is that the Thread pools worker threads
have to sequentially execute their tasks one after the other
and the thread sequence for the 10 have to be maintained...
One simple way to solve this is use below state in runnable class
private final int index;
private final AtomicInteger atomicInteger;
private final CyclicBarrier cyclicBarrier;
index - is responsible for conditional verification i.e., which number this thread should print.
atomicInteger - shared across all threads for current number.
Cyclic barrier - makes all threads to wait unit a every thread completes a cycle/iteration.
Code sample:
public class PrintSequence {
public static void main(String[] args) {
ExecutorService executorService = Executors.newFixedThreadPool(10);
final AtomicInteger atomicInteger = new AtomicInteger(1);
final CyclicBarrier cyclicBarrier = new CyclicBarrier(10, ()-> {
System.out.println("a cycle done");
});
IntStream.rangeClosed(0, 9)
.boxed()
.map(i -> new PrintSequenceTask(i, atomicInteger, cyclicBarrier))
.map(p -> executorService.submit(p))
.collect(Collectors.toList());
executorService.shutdown();
}
}
class PrintSequenceTask implements Runnable {
private final int index;
private final AtomicInteger atomicInteger;
private final CyclicBarrier cyclicBarrier;
PrintSequenceTask(int index, AtomicInteger atomicInteger, CyclicBarrier cyclicBarrier) {
this.index = index;
this.atomicInteger = atomicInteger;
this.cyclicBarrier = cyclicBarrier;
}
#Override
public void run(){
for(int i=1; i<10;i++){
while (((atomicInteger.get()-index-1)%10 != 0)){}
System.out.println(Thread.currentThread().getName()+" "+(atomicInteger.get()));
atomicInteger.getAndIncrement();
await();
}
}
public void await(){
try {
cyclicBarrier.await();
} catch (InterruptedException e) {
e.printStackTrace();
} catch (BrokenBarrierException e) {
e.printStackTrace();
}
}
}
public class BigSequence {
public static void main(String[] args) {
BigPrintNum p = new BigPrintNum();
int max = 20;
int no_threads = 11;
for(int i=0;i<no_threads;i++){
boolean b[] = new boolean[no_threads];
b[i] = true;
Thread t = new Thread(new BigPrint(p, max, b,no_threads));
t.start();
}
}
}
class BigPrint implements Runnable {
int num=0;
BigPrintNum p;
int max;
int no_threads;
boolean b[];
public BigPrint(BigPrintNum p,int max,boolean b[],int no_threads){
this.p = p;
this.max = max;
this.b = b;
this.no_threads = no_threads;
}
#Override
public void run() {
int n = 0;
for(int i=0;i<no_threads;i++){
if(b[i] == true){
n = i;
num = i;
}
}
while(num<=max){
p.print(num, n, no_threads);
num += no_threads;
}
}
}
class BigPrintNum {
int turn = 0;
public synchronized void print(int n,int i,int no_threads){
while(this.turn != i){
try{
wait();
}catch(InterruptedException e){
e.printStackTrace();
}
}
System.out.println(i + "th seq = " + n);
this.turn = (i+1)%no_threads;
notifyAll();
}
}
Its a generic one, where we can use any number of threads and use any max value.
public class ThreadSequence
{
public static int totalThread;
public static void main(String[] args)
{
MyLock myLock = new MyLock();
totalThread = 10;
for(int i=1;i<=totalThread;i++)
{
MyThread myThread = new MyThread(i,myLock);
myThread.start();
}
}
}
class MyLock
{
public int counter = 0;
}
MyThread Class
class MyThread extends Thread{
public MyLock lock;
public int no;
public MyThread(int no,MyLock lock)
{
super("My Thread No "+no);
this.no = no;
this.lock = lock;
}
public void run()
{
synchronized (lock)
{
while(true)
{
while(lock.counter%ThreadSequence.totalThread !=(this.no-1))
{
try
{
if(lock.counter > 99)
{
break;
}
lock.wait();
} catch (InterruptedException e)
{
e.printStackTrace();
}
}
if(lock.counter > 99)
{
break;
}
System.out.println("Current Thread "+Thread.currentThread().currentThread()+" --- Current Count "+(lock.counter+1));
lock.counter = lock.counter +1 ;
lock.notifyAll();
}
}
}
}
print 1 to 100 number alternatively by each thread similar way you can print for 10 threads- m1 and m2 like
m1-1
m2-2
m3-3
m4-4
public class MultiThread extends Thread {
static volatile int num=0;
public static void main(String[] args) {
MultiThread m1= new MultiThread();
MultiThread m2= new MultiThread();
m1.setName("m1");
m1.setPriority(5);
m2.setName("m2");
m2.setPriority(5);
m1.start();
m2.start();
}
#Override
public void run() {
while(num<100) {
num +=1;
print();
}
}
private void print(){
synchronized(this) {
System.out.println(currentThread().getName()+" "+ num);
try {
currentThread().wait(500);
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
}
}
}
The simple thing to do is to hold common resource for all of them.
Hold a List and every thread will insert into the list, in the end you can sort and print..
If you want them to do it on your order it won't be very effective because you won't need 10 threads to do it..
This way it will be faster and will use 10 threads to do some work, but when everyone finish you still need to do some work
public class PrintNumbersbyThreads implements Runnable {
private int i;
public PrintNumbersbyThreads(int i) {
this.i = i;
}
public static void main(String[] args) {
PrintNumbersbyThreads p = new PrintNumbersbyThreads(1);
PrintNumbersbyThreads p2 = new PrintNumbersbyThreads(2);
PrintNumbersbyThreads p3 = new PrintNumbersbyThreads(3);
Thread t1 = new Thread(p, "t1");
Thread t2 = new Thread(p2, "t2");
Thread t3 = new Thread(p3, "t3");
t1.start();
try {
t1.join();
t2.start();
t2.join();
t3.start();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
#Override
public void run() {
System.out.println("\n" + Thread.currentThread().getName() + " prints ");
for (int j = 0; j < 10; j++) {
System.out.print(i + " ");
i = i + 10;
}
}
}
Written sample code 3 Threads and the output is
t1 prints:
1 11 21 31 41 51 61 71 81 91
t2 prints:
2 12 22 32 42 52 62 72 82 92
t3 prints:
3 13 23 33 43 53 63 73 83 93
Hope this is what you are Looking for?
I have written one generic code which will take the number till where you want to print and the number of threads to be used.
public class ThreadedPrinting {
private Object locks[];
private static class Printer extends Thread {
int curVal;
int endVal;
Object myLock;
Object nextLock;
int step;
public Printer(int startFrom, int endVal, int step, Object myLock, Object nextLock){
this.curVal = startFrom;
this.endVal = endVal;
this.step = step;
this.myLock = myLock;
this.nextLock = nextLock;
this.step = step;
}
#Override
public void run(){
synchronized(myLock) {
while (curVal <= endVal) {
try {
myLock.wait();
System.out.println(curVal);
curVal += step;
}
catch(InterruptedException e) {}
synchronized(nextLock) {
nextLock.notify();
}
}
}
synchronized(nextLock) {
nextLock.notify(); /// this ensures all worker threads exiting at the end
}
}
} // Printer
public ThreadedPrinting(int maxNum, int threads) {
locks = new Object[threads];
int i;
for(i = 0; i < threads; ++i) locks[i] = new Object();
for(i = 0; i < threads -1 ; ++i) {
Printer curPrinter = new Printer(i, maxNum, threads, locks[i], locks[i+1]);
curPrinter.start();
}
Printer lastPrinter = new Printer(i, maxNum, threads, locks[threads - 1], locks[0]);
lastPrinter.start();
}
public void start() {
synchronized (locks[0]) {
locks[0].notify();
}
}
public static void main(String[] args) {
ThreadedPrinting printer = new ThreadedPrinting(1000,7);
printer.start();
}
}
The same problem can be solved by usign Phaser as well but the order is not restrictive but will be in round-robin fashion. I have provided the solution for similar problem here.