Java threadsafe counter out of bound - java

I want to implement simple threadsafe counter. The numbers are in the right order so that part is ok, the only problem is the condition is not always met and sometimes the numbers go up to 51 or 52.
Should I use the tag synchronized also around the while loop?
I mean, I can double check and put a condition in the method printAndIncrement but that doesn't seem very elegant.
public class MyCounter implements Runnable {
static int currentValue = 0;
private static synchronized void printAndIncrement() {
System.out.print(Thread.currentThread().getName() + ": " + currentValue + "\n");
currentValue++;
}
#Override
public void run() {
while (currentValue <= 50) {
printAndIncrement();
}
}
public static void main(String[] args) {
MyCounter counter = new MyCounter();
Thread thread1 = new Thread(counter);
Thread thread2 = new Thread(counter);
Thread thread3 = new Thread(counter);
thread1.start();
thread2.start();
thread3.start();
}
}

The check currentValue <= 50 and the call to printAndIncrement must be in the same synchronized block. Otherwise this problem is going to happen.
Let currentValue be 50. All three threads can do the check that the current value is no more than 50 and then try to call printAndIncrement(); simultaneously.
Due to the synchronized void printAndIncrement() the threads will execute this method sequentially, but for the first thread the currentValue will be 50, for the second thread it will be 51 and for the third thread it will be 52.

The problem is that your boundary check and the increment are not both synchronized, which defeats the point of synchronization altogether.
The best alternative I can suggest that allows both synchronizing read/update and allowing your loop to stop would be to make the incrementing method return a boolean:
/** Prints and increments, returning true if max value has not been reached */
private static synchronized boolean printAndIncrement() {
if(currentValue < 51) {
System.out.print(Thread.currentThread().getName()
+ ": " + currentValue + "\n");
currentValue++;
return true;
} else {
return false;
}
}
And change the run method to:
public void run() {
while (printAndIncrement()) {
//nothing needs to be done here
}
}

Related

What is happening while printing even odd numbers using interrupt method?

I am trying to print even odd numbers using two threads with interrupt method.
I refereed code from internet and wrote a code showing below.It prints properly but after prints 20,program is continuing it's execution.
What change do i have to make in the code to stop the execution of the program?
Without oldNum check code is working fine. Is there any logic to provide oldNum check ?
If I remove Thread.sleep(1000L) from Line-a then it only prints "Even Thread prints 20" and continue execution.What is happening here?
Provided break points inside run() method and inside for loop of main method ,run() methods break points are not hitting.Why this is happening?
In short I want to know what is the code flow here.
Thanks
Vikash
public class PrintOddEvenUsingInterrupt {
public static volatile int count;
public static void main(String[] args) throws InterruptedException {
Thread oddThread = new Thread(new OddInterruptThread(), "Odd Thread ");
Thread evenThread = new Thread(new EvenInterruptThread(),"Even Thread ");
oddThread.start();
evenThread.start();
for (int i = 0; i < 20; i++) {
count++;
oddThread.interrupt();//Break points works here
evenThread.interrupt();
Thread.sleep(1000L);// Line-a
}
}
static class OddInterruptThread implements Runnable {
public void run() {
int oldNum = 0;//Break points doesn't works here
while (true) {
try {
Thread.sleep(Integer.MAX_VALUE);
} catch (InterruptedException e) {
}
if (oldNum != count && count % 2 == 1) {
System.out.println(Thread.currentThread().getName()
+ " prints " + count);
oldNum = count;
}
}
}
}
static class EvenInterruptThread implements Runnable {
public void run() {
int oldNum = 0;//Break points doesn't works here
while (true) {
try {
Thread.sleep(Integer.MAX_VALUE);
} catch (InterruptedException e) {
}
if (oldNum != count && count % 2 == 0) {
System.out.println(Thread.currentThread().getName()
+ " prints " + count);
oldNum = count;
}
}
}
}
}
The reason your program is not stopping is: while your main thread exits, your odd and even threads sleeps in infinite loop.
You will need to define a stopping condition for your threads to come out.
One way to achieve this is via using conditions.
Eg:
public volatile static boolean oddFinished = false;
public volatile static boolean evenFinished = false;
Then in your threads, instead of looping infinitely, loop against condition
while (! oddFinished){
// also change your thread sleep to sleep for fewer time interval (say 1000L or whatever your program wants to wait for)
}
Do the same for even thread...
while (! evenFinished){
// also change your thread sleep to sleep for fewer time interval (say 1000L or whatever your program wants to wait for)
}
And in the main thread, you can add the following code after your for loop ends...
oddFinished = true;
evenFinished = true;
oddThread.join();
evenThread.join();
This will allow your code to stop gracefully.
I think the simplest solution will be to make your threads demons.
Just add the following lines before starting your thteads.
oddThread.setDaemon(true);
evenThread.setDaemon(true);
And your program will exit immediately after exiting from main.

counter value is getting rested in multithreading

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.

Notify seems to be waking up more than one thread

I am doing a sample program with wait() and notify(), but when notify() is called, more than one thread is wakes up instead of one.
The code is:
public class MyQueue<T> {
Object[] entryArr;
private volatile int addIndex;
private volatile int pending = -1;
private final Object lock = new Object();
private volatile long notifiedThreadId;
private int capacity;
public MyQueue(int capacity) {
entryArr = new Object[capacity];
this.capacity = capacity;
}
public void add(T t) {
synchronized (lock) {
if (pending >= 0) {
try {
pending++;
lock.wait();
System.out.println(notifiedThreadId + ":" + Thread.currentThread().getId());
} catch (InterruptedException e) {
e.printStackTrace();
}
} else if (pending == -1) {
pending++;
}
}
if (addIndex == capacity) { // its ok to replace existing value
addIndex = 0;
}
try {
entryArr[addIndex] = t;
} catch (ArrayIndexOutOfBoundsException e) {
System.out.println("ARRAYException:" + Thread.currentThread().getId() + ":" + pending + ":" + addIndex);
e.printStackTrace();
}
addIndex++;
synchronized (lock) {
if (pending > 0) {
pending--;
notifiedThreadId = Thread.currentThread().getId();
lock.notify();
} else if (pending == 0) {
pending--;
}
}
}
}
public class TestMyQueue {
public static void main(String args[]) {
final MyQueue<String> queue = new MyQueue<>(2);
for (int i = 0; i < 200; i++) {
Runnable r = new Runnable() {
#Override
public void run() {
for (int i = 0; i < Integer.MAX_VALUE; i++) {
queue.add(Thread.currentThread().getName() + ":" + i);
}
}
};
Thread t = new Thread(r);
t.start();
}
}
}
After some time, I see two threads being wake up by single thread. The output looks like:
91:114
114:124
124:198
198:106
106:202
202:121
121:40
40:42
42:83
83:81
81:17
17:189
189:73
73:66
66:95
95:199
199:68
68:201
201:70
70:110
110:204
204:171
171:87
87:64
64:205
205:115
Here I see 115 thread notified two threads, and 84 thread notified two threads; because of this we are seeing the ArrayIndexOutOfBoundsException.
115:84
115:111
84:203
84:200
ARRAYException:200:199:3
ARRAYException:203:199:3
What is the issue in the program?
What is the issue in the program?
You have a couple of problems with your code that may be causing this behavior. First, as #Holder commented on, there are a lot of code segments that can be run by multiple threads simultaneously that should be protected using synchronized blocks.
For example:
if (addIndex == capacity) {
addIndex = 0;
}
If multiple threads run this then multiple threads might see addIndex == capacity and multiple would be overwriting the 0th index. Another example is:
addIndex++;
This is a classic race condition if 2 threads try to execute this statement at the same time. If addIndex was 0 beforehand, after the 2 threads execute this statement, the value of addIndex might be 1 or 2 depending on the race conditions.
Any statements that could be executed at the same time by multiple threads have to be properly locked within a synchronized block or otherwise protected. Even though you have volatile fields, there can still be race conditions because there are multiple operations being executed.
Also, a classic mistake is to use if statements when checking for over or under flows on your array. They should be while statements to make sure you don't have the class consumer producer race conditions. See my docs here or take a look at the associated SO question: Why does java.util.concurrent.ArrayBlockingQueue use 'while' loops instead of 'if' around calls to await()?

Error in Two Process critical section solution

I have applied two process critical section solution to two threads instead of processes. My code is:
class Main
{
static boolean flag[];
static int turn;
static int count;
synchronized static void print(char ch,int n)
{
int i;
System.out.println(ch);
for(i=0;i<n;i++){
System.out.println(i);
}
}
public static void main(String[] args) throws IOException
{
flag = new boolean[2];
flag[0] = flag[1] = false;
turn = 0;
count = 0;
ThreadLevelOne t1 = new ThreadLevelOne('a');
ThreadLevelTwo t2 = new ThreadLevelTwo('b');
t1.start();
t2.start();
}
static class ThreadLevelOne extends Thread{
private char ch;
public ThreadLevelOne(char ch){
this.ch = ch;
}
public void run(){
while(true)
{
flag[0] = true;
turn = 1;
while(flag[1] && turn == 1);
print(ch,3);
count++;
System.out.println("Counter is : " + count);
flag[0] = false;
}
}
}
static class ThreadLevelTwo extends Thread{
private char ch;
public ThreadLevelTwo(char ch){
this.ch = ch;
}
public void run()
{
while(true)
{
flag[1] = true;
turn = 0;
while(flag[0] && turn == 0);
print( ch, 4);
count++;
System.out.println("Counter is : " + count);
flag[1] = false;
}
}
}
}
On executing the above code, it does not run infinitely but halts at arbitrary counter value on each execution. Is this a valid application of the two process solution to threads? If yes, then why is program halting at arbitrary counter value? If no, then how can this be achieved in threads?
Edit after the answer of codeBlind:
output: Program execution halts at this stage
Even if i dont increment the counter value, then also the program halts after a certain time
You're a victim of concurrently executing non-atomic operations, specifically count++, as well as the way you are using flags in each thread. But for simplicity's sake, let's talk about count++. The ++ operator actually executes three commands, each in their own clock-cycle:
read value of count
add 1 to value retrieved from count
store new value into count
The problem you're seeing is a result of these commands being interleaved across two threads. Thread A may not have stored the new count value by the time that Thread B attempts to read it.
A quick fix would be to use AtomicInteger for count instead of primitive int - AtomicInteger guarantees thread safety for integer operations.
EDIT
There are other race conditions in this code as well. Each thread's while loop argument (e.g. flag[0] && turn == 0) is non-atomic, but both threads are capable of modifying turn. You've left open the possibility that one thread could set turn before the other thread's while argument is fully evaluated, causing your threads to deadlock down the road.
If you only wish to guarantee that each thread must not be inside the while loop while the other thread is, then you should instead write each of your while loops to look something like this:
while(true){
synchronized(Main.class){
print( ch, 4);
count++;
System.out.println("Counter is : " + count);
}
}
If you want to guarantee that each thread must "take turns", you should look into using wait() and notify().
Ok so I figured it out, the issue is that each thread needs to pause in order for the other thread to run.
Instead of just spinning the cpu using:
while(flag[0] && turn == 0);
You need to pause the thread by calling the sleep method.
while(flag[0] && turn == 0){
try {
this.sleep(100);
} catch (InterruptedException e) {
e.printStackTrace();
}
}

Thread Scheduling - Shared Array

I need two threads to write one a shared array of ints. Both threads need to write on all the elements of that array. Each thread will write either 1 or 7, and the result should be like 171717171 (or 71717171). To do that I have the first Thread1 write at position 0, then wait. Thread2 now writes at position 0 and 1, notifies Thread1, and waits. Thread1 writes at position 1 and 2, notifies Thread2 and waits, etc. With the following code I get correct output, although when run with JPF it finds a deadlock. Its become really frustrating since I can not find whats wrong with it. Any advice would be appreciated.
import java.util.logging.Level;
import java.util.logging.Logger;
public class WriterThreadManager {
private int[] array = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
private Thread thread7;
private Thread thread1;
public static void main(String[] args) {
WriterThreadManager mng = new WriterThreadManager();
mng.exec();
}
public WriterThreadManager() {
thread7 = new Thread(new WriterRunnable(this, 7));
thread1 = new Thread(new WriterRunnable(this, 1));
}
public void overwriteArray(int pos, int num) {
array[pos] = num;
printArray();
}
private void printArray() {
for (int i = 0; i < array.length; i++) {
System.out.print(array[i]);
}
System.out.println("");
}
public synchronized void stopThread() {
try {
this.wait();
} catch (InterruptedException ex) {
Logger.getLogger(WriterThreadManager.class.getName()).log(Level.SEVERE, null, ex);
}
}
public synchronized void wakeUpThread() {
notifyAll();
}
private void exec() {
thread7.start();
thread1.start();
}
public int length() {
return array.length;
}
}
public class WriterRunnable implements Runnable {
private WriterThreadManager mng;
private int numberToWrite;
private static boolean flag = true;
#Override
public void run() {
int counter = 0;
int j = 0;
//first thread to get in should write only at
//position 0 and then wait.
synchronized (mng) {
if (flag) {
flag = false;
mng.overwriteArray(0, numberToWrite);
j = 1;
waitForOtherThread();
}
}
for (int i = j; i < mng.length(); i++) {
mng.overwriteArray(i, numberToWrite);
counter++;
if (i == mng.length() - 1) {
mng.wakeUpThread();
break;
}
if (counter == 2) {
waitForOtherThread();
counter = 0;
}
}
}
private void waitForOtherThread() {
mng.wakeUpThread();
mng.stopThread();
}
public WriterRunnable(WriterThreadManager ar, int num) {
mng = ar;
numberToWrite = num;
}
}
p.s: an example of the execution:
1000000000
7000000000
7700000000
7100000000
7110000000
7170000000
7177000000
7171000000
7171100000
7171700000
7171770000
7171710000
7171711000
7171717000
7171717700
7171717100
7171717110
7171717170
7171717177
7171717171
The error snapshot from JPF is the following:
thread java.lang.Thread:{id:1,name:Thread-1,status:WAITING,priority:5,lockCount:1,suspendCount:0}
waiting on: WriterThreadManager#152
call stack:
at java.lang.Object.wait(Object.java)
at WriterThreadManager.stopThread(WriterThreadManager.java:43)
at WriterRunnable.waitForOtherThread(WriterRunnable.java:53)
at WriterRunnable.run(WriterRunnable.java:45)
thread java.lang.Thread:{id:2,name:Thread-2,status:WAITING,priority:5,lockCount:1,suspendCount:0}
waiting on: WriterThreadManager#152
call stack:
at java.lang.Object.wait(Object.java)
at WriterThreadManager.stopThread(WriterThreadManager.java:43)
at WriterRunnable.waitForOtherThread(WriterRunnable.java:53)
at WriterRunnable.run(WriterRunnable.java:45)
I believe the race is due to this method:
private void waitForOtherThread() {
mng.wakeUpThread();
mng.stopThread();
}
While the individual wakeUpThread() and stopThread() methods are synchronized, you have the opportunity for unexpected thread scheduling between these calls.
Consider:
thread7 - notify thread1 to wakup
thread1 - wake up
thread1 - work to completion
thread1 - notify thread7 to wakeup
thread1 - wait to be notified to wakeup
thread7 - wait to be notified to wakeup
In this case you have deadlocked because thread1 sent its notifyAll() before thread7 had a chance to wait() for it.
Running in a different context can mess with your timing and cause these types of behaviors to appear.
To avoid this I suggestion doing this:
private void waitForOtherThread() {
synchronized(mng) {
mng.wakeUpThread();
mng.stopThread();
}
}
Or better yet, use a semaphore as #KumarVivekMitra suggested. Semaphores combine both the notification system and a counter so that the order of the notify and wait don't matter.
- I think a better approach here would be java.util.Semaphores, which will help you to decide the access over the objects resources by specific numbers of threads at a time.
- Well you can also use the SingleThreadExecutor to solve this, which starts and completes a task before moving on to the 2nd task, so there will be No need of synchronization needed here from your side.
I don't think you need any sort of coordination here. Just have one thread write the even locations and the other thread write the odd locations. Let them both go as fast as they can. Done!

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