I have written a short program in order to check the effect of the race condition. Class Counter is given below. The class has two methods to update the counter instance variable c. On purpose, I added a random code in both methods , see related code variable i, to increase the probability of their interleaved execution when accessed by two threads.
In the main() method of my program, I put in a loop the following code
t1=new Thread() { public void run(){objCounter.increment();}};
t2=new Thread() { public void run(){objCounter.decrement();}};
t1.start();
t2.start();
try{
t1.join();
t2.join();
}
catch (InterruptedException IE) {}
Then I printed the different values of c in the objCount... Further to the expected values 1, 0, -1, the program displays also the unexpected values: -2,-1, -3, even 4
I sincerely can't see what threads interleaving will lead to the unexpected values given above. Ideally, I should look at the assembly code to see how the statements c++, and c-- got translated...regardless, I Think there is another reason behind the unexpected values.
class Counter{
private volatile int c=0;
public void increment(){
int i=9;
i=i+7;
c++;
i=i+3;
}
public void decrement() {
int i=9;
i=i+7;
c--;
i=i+3;
}
public int value(){ return c; }
}
Even if you marked an int as volatile, that kind of operations are not atomic. Try to replace your primitive int with a Thread Safe Class like:
https://docs.oracle.com/javase/8/docs/api/java/util/concurrent/atomic/AtomicInteger.html
Or just access it through a synchronyzed method.
I put in a loop the following code
You don't show reinitialization of the objCounter variable; this suggests that you're reusing the variable between loop iterations.
As such, you can get -2 from the situation resulting in -1 (e.g. Thread 1 read, Thread 2 read, T1 write, T2 write) happening twice.
In order to avoid reusing the state from previous runs, you should declare and initialize the objCounter variable inside the loop:
for (...) {
Counter objCounter = new Counter();
t1=new Thread() { public void run(){objCounter.increment();}};
t2=new Thread() { public void run(){objCounter.decrement();}};
// ... Start/join the threads.
}
It can't be declared before the loop and initialized inside the loop, because then it is not effectively final, which is required (that, or actual finality) to refer to it inside the anonymous classes of the threads.
On purpose, I added a random code in both methods , see related code variable i, to increase the probability of their interleaved execution when accessed by two threads.
As an aside, this your random code does nothing of the sort.
There is no requirement for Java to execute the statements in program order, only to appear to execute them in the program order from the perspective of the current thread.
These statements may be executed before or after the c++/--, if they are executed at all - they could simply be detected as useless.
You may as well just remove this code, it really only serves to obfuscate.
Related
I'm learning multithreading. Can anyone tell why here the output is always 100, even though there are two threads which are doing 100 increments?
public class App {
public static int counter = 0;
public static void process() {
Thread thread1 = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < 100; ++i) {
++counter;
}
}
});
Thread thread2 = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < 100; ++i) {
++counter;
}
}
});
thread1.start();
thread2.start();
}
public static void main(String[] args) {
process();
System.out.println(counter);
}
}
The output is 100.
You're only starting the threads, not waiting for them to complete before you print the result. When I run your code, the output is 0, not 100.
You can wait for the threads with
thread1.join();
thread2.join();
(at the end of the process() method). When I add those, I get 200 as output. (Note that Thread.join() throws an InterruptedException, so you have to catch or declare this exception.)
But I'm 'lucky' to get 200 as output, since the actual behaviour is undefined as Stephen C notes. The reason why is one of the main pitfalls of multithreading: your code is not thread safe.
Basically: ++counter is shorthand for
read the value of counter
add 1
write the value of counter
If thread B does step 1 while thread A hasn't finished step 3 yet, it will try to write the same result as thread A, so you'll miss an increment.
One of the ways to solve this is using AtomicInteger, e.g.
public static AtomicInteger counter = new AtomicInteger(0);
...
Thread thread1 = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < 100; ++i) {
counter.incrementAndGet();
}
}
});
Can anyone tell why here the output is always 100, even though there are two threads which are doing 100 increments?
The reason is that you have two threads writing a shared variable and a third reading, all without any synchronization. According to the Java Memory Model, this means that the actual behavior of your example is unspecified.
In reality, your main thread is (probably) printing the output before the second thread starts. (And apparently on some platforms, it prints it before the first one starts. Or maybe, it is seeing a stale value for counter. It is a bit hard to tell. But this is all within the meaning of unspecified)
Apparently, adding join calls before printing the results appears to fix the problem, but I think that is really by luck1. If you changed 100 to a large enough number, I suspect that you would find that incorrect counter values would be printed once again.
Another answer suggests using volatile. This isn't a solution. While a read operation following a write operation on a volatile is guaranteed to give the latest value written, that value may be a value written by another thread. In fact the counter++ expression is an atomic read followed by an atomic write ... but the sequence is not always atomic. If two or more threads do this simultaneously on the same variable, they are liable to lose increments.
The correct solutions to this are to either using an AtomicInteger, or to perform the counter++ operations inside a synchronized block; e.g.
for (int i = 0; i < 100; ++i) {
synchronized(App.class) {
++counter;
}
}
Then it makes no difference that the two threads may or may not be executed in parallel.
1 - What I think happens is that the first thread finishes before the second thread starts. Starting a new thread takes a significant length of time.
In Your case, There are three threads are going to execute: one main, thread1 and thread2. All these three threads are not synchronised and in this case Poor counter variable behaviour will not be specific and particular.
These kind of Problem called as Race Conditions.
Case1: If i add only one simple print statement before counter print like:
process();
System.out.println("counter value:");
System.out.println(counter);
in this situation scenario will be different. and there are lot more..
So in these type of cases, according to your requirement modification will happen.
If you want to execute one thread at time go for Thread join like:
thread1.join();
thread2.join();
join() is a Thread class method and non static method so it will always apply on thread object so apply join after thread start.
If you want to read about Multi threading in java please follow; https://docs.oracle.com/cd/E19455-01/806-5257/6je9h032e/index.html
You are checking the result before threads are done.
thread1.start();
thread2.start();
try{
thread1.join();
thread2.join();
}
catch(InterruptedException e){}
And make counter variable volatile.
I am trying to see how volatile works here. If I declare cc as volatile, I get the output below. I know thread execution output varies from time to time, but I read somewhere that volatile is the same as synchronized, so why do I get this output? And if I use two instances of Thread1 does that matter?
2Thread-0
2Thread-1
4Thread-1
3Thread-0
5Thread-1
6Thread-0
7Thread-1
8Thread-0
9Thread-1
10Thread-0
11Thread-1
12Thread-0
public class Volexample {
int cc=0;
public static void main(String[] args) {
Volexample ve=new Volexample();
CountClass count =ve.new CountClass();
Thread1 t1=ve.new Thread1(count);
Thread2 t2=ve.new Thread2(count);
t1.start();
t2.start();
}
class Thread1 extends Thread{
CountClass count =new CountClass();
Thread1(CountClass count ){
this.count=count;
}
#Override
public void run() {
/*for(int i=0;i<=5;i++)
count.countUp();*/
for(int i=0;i<=5;i++){
cc++;
System.out.println(cc + Thread.currentThread().getName());
}
}
}
class Thread2 extends Thread {
CountClass count =new CountClass();
Thread2(CountClass count ){
this.count=count;
}
#Override
public void run(){
/*for(int i=0;i<=5;i++)
count.countUp();*/
for(int i=0;i<=5;i++){
cc++;
System.out.println(cc + Thread.currentThread().getName());
}
}
}
class CountClass{
volatile int count=0;
void countUp(){
count++;
System.out.println(count + Thread.currentThread().getName());
}
}
}
In Java, the semantics of the volatile keyword are very well defined. They ensure that other threads will see the latest changes to a variable. But they do not make read-modify-write operations atomic.
So, if i is volatile and you do i++, you are guaranteed to read the latest value of i and you are guaranteed that other threads will see your write to i immediately, but you are not guaranteed that two threads won't interleave their read/modify/write operations so that the two increments have the effect of a single increment.
Suppose i is a volatile integer whose value was initialized to zero, no writes have occurred other than that yet, and two threads do i++;, the following can happen:
The first thread reads a zero, the latest value of i.
The second threads reads a zero, also the latest value of i.
The first thread increments the zero it read, getting one.
The second thread increments the zero it read, also getting one.
The first thread writes the one it computed to i.
The second thread writes the one it computed to i.
The latest value written to i is one, so any thread that accesses i now will see one.
Notice that an increment was lost, even though every thread always read the latest value written by any other thread. The volatile keyword gives visibility, not atomicity.
You can use synchronized to form complex atomic operations. If you just need simple ones, you can use the various Atomic* classes that Java provides.
A use-case for using volatile would be reading/writing from memory that is mapped to device registers, for example on a micro-controller where something other than the CPU would be reading/writing values to that "memory" address and so the compiler should not optimise that variable away .
The Java volatile keyword is used to mark a Java variable as "being stored in main memory". That means, that every read of a volatile variable will be read from the computer's main memory, and not from the cache and that every write to a volatile variable will be written to main memory, and not just to the cache.
It guarantees that you are accessing the newest value of this variable.
P. S. Use larger loops to notice bugs. For example try to iterate 10e9 times.
I'm trying to learn about threads and I do not understand the join() method.
I have a Thread (ThreadAdd.java) which adds 1 to a static int.
public class ThreadAdd extends Thread{
public static int count;
#Override
public void run() {
try {
Thread.sleep(100);
} catch (InterruptedException ex) {
Logger.getLogger(ThreadAdd.class.getName()).log(Level.SEVERE, null, ex);
}
ThreadAdd.count++;
}
}
In my main method I launch 2 threads :
public static void main(String[] args) throws InterruptedException {
ThreadAdd s1 = new ThreadAdd();
ThreadAdd s2 = new ThreadAdd();
s1.start();s2.start();
s1.join();
s2.join();
System.out.println(ThreadAdd.count);
}
I do not understand why most of the time the result is 2 but sometimes it returns 1.
The reason why you sometimes see 1 is not because join() fails to wait for the thread to finish, but because both threads tried to modify the value concurrently. When this happens, you may see unexpected results: for example, when both threads try to increment count which is zero, they both could read zero, then add 1 to it, and store the result. Both of them will store the same exact result, i.e. 1, so that's what you are going to see no matter how long you wait.
To fix this problem, add synchronized around the increment, or use AtomicInteger:
public static AtomicInteger count = new AtomicInteger(0);
#Override
public void run() {
try {
Thread.sleep(100);
} catch (InterruptedException ex) {
Logger.getLogger(ThreadAdd.class.getName()).log(Level.SEVERE, null, ex);
}
ThreadAdd.count.incrementAndGet();
}
The join method is not the real issue here. The problem is that your counter is not prepared for interthread synchronization, which may lead to each thread observing a different value in count.
It is highly recommended that you study some topics of concurrent programming, including how it is handled in Java.
Because you're not synchronizing the increment of the integer count. The two threads may interleave while incrementing the variable.
See http://docs.oracle.com/javase/tutorial/essential/concurrency/interfere.html for an explanation. The example in the link is similar to your example and a solution provided to avoid this thread interference is to use atomic variables like java.util.concurrent.atomic.AtomicInteger.
Your count variable isn't volatile, and so there's no requirement for threads to check its value each time, and occasionally instruction ordering will cause errors like that.
In fact, though, since count++ is syntactic sugar for count = count + 1, even making the variable volatile won't ensure that you don't have the problem, since there's a race condition between the read and the subsequent write.
To make code like this safe, use an AtomicInteger instead.
This has nothing to do with the join. The thread that waits by using join() is your main thread. The two other threads are not waiting for anything. And the join is not causing them to do anything differently.
And as the other answers said, the two threads are concurrently writing to the same variable, and therefore you get the result you see.
Perhaps you were expecting the join() to delay one of the threads so that it doesn't work concurrently with the other, but that's not how it works. The only thread that is delayed is the caller of join(), not the target thread.
class Counter
{
public int i=0;
public void increment()
{
i++;
System.out.println("i is "+i);
System.out.println("i/=2 executing");
i=i+22;
System.out.println("i is (after i+22) "+i);
System.out.println("i+=1 executing");
i++;
System.out.println("i is (after i++) "+i);
}
public void decrement()
{
i--;
System.out.println("i is "+i);
System.out.println("i*=2 executing");
i=i*2;
System.out.println("i is after i*2"+i);
System.out.println("i-=1 executing");
i=i-1;
System.out.println("i is after i-1 "+i);
}
public int value()
{
return i;
} }
class ThreadA
{
public ThreadA(final Counter c)
{
new Thread(new Runnable(){
public void run()
{
System.out.println("Thread A trying to increment");
c.increment();
System.out.println("Increment completed "+c.i);
}
}).start();
}
}
class ThreadB
{
public ThreadB(final Counter c)
{
new Thread(new Runnable(){
public void run()
{
System.out.println("Thread B trying to decrement");
c.decrement();
System.out.println("Decrement completed "+c.i);
}
}).start();
}
}
class ThreadInterference
{
public static void main(String args[]) throws Exception
{
Counter c=new Counter();
new ThreadA(c);
new ThreadB(c);
}
}
In the above code, ThreadA first got access to Counter object and incremented the value along with performing some extra operations. For the very first time ThreadA does not have a cached value of i. But after the execution of i++ (in first line) it will get cache the value. Later on the value is updated and gets 24. According to the program, as the variable i is not volatile so the changes will be done in the local cache of ThreadA,
Now when ThreadB accesses the decrement() method the value of i is as updated by ThreadA i.e. 24. How could that be possible?
Assuming that threads won't see each updates that other threads make to shared data is as inappropriate as assuming that all threads will see each other's updates immediately.
The important thing is to take account of the possibility of not seeing updates - not to rely on it.
There's another issue besides not seeing the update from other threads, mind you - all of your operations act in a "read, modify, write" sense... if another thread modifies the value after you've read it, you'll basically ignore it.
So for example, suppose i is 5 when we reach this line:
i = i * 2;
... but half way through it, another thread modifies it to be 4.
That line can be thought of as:
int tmp = i;
tmp = tmp * 2;
i = tmp;
If the second thread changes i to 4 after the first line in the "expanded" version, then even if i is volatile the write of 4 will still be effectively lost - because by that point, tmp is 5, it will be doubled to 10, and then 10 will be written out.
As specified in JLS 8.3.1.4:
The Java programming language allows threads to access shared
variables (ยง17.1). As a rule, to ensure that shared variables are
consistently and reliably updated, a thread should ensure that it has
exclusive use of such variables by obtaining a lock that,
conventionally, enforces mutual exclusion for those shared variables........A field may be
declared volatile, in which case the Java Memory Model ensures that
all threads see a consistent value for the variable
Although not always but there is still a chance that the shared values among threads are not consistenly and reliably updated, which would lead to some unpredictable outcome of program. In code given below
class Test {
static int i = 0, j = 0;
static void one() { i++; j++; }
static void two() {
System.out.println("i=" + i + " j=" + j);
}
}
If, one thread repeatedly calls the method one (but no more than Integer.MAX_VALUE times in all), and another thread repeatedly calls the method two then method two could occasionally print a value for j that is greater than the value of i, because the example includes no synchronization and, the shared values of i and j might be updated out of order.
But if you declare i and j to be volatile , This allows method one and method two to be executed concurrently, but guarantees that accesses to the shared values for i and j occur exactly as many times, and in exactly the same order, as they appear to occur during execution of the program text by each thread. Therefore, the shared value for j is never greater than that for i,because each update to i must be reflected in the shared value for i before the update to j occurs.
Now i came to know that common objects (the objects that are being shared by multiple threads) are not cached by those threads. As the object is common, Java Memory Model is smart enough to identify that common objects when cached by threads could produce surprising results.
How could that be possible?
Because there is nowhere in the JLS that says values have to be cached within a thread.
This is what the spec does say:
If you have a non-volatile variable x, and it's updated by a thread T1, there is no guarantee that T2 can ever observe the change of x by T1. The only way to guarantee that T2 sees a change of T1 is with a happens-before relationship.
It just so happens that some implementations of Java cache non-volatile variables within a thread in certain cases. In other words, you can't rely on a non-volatile variable being cached.
I have experience this weird behavior of volatile keyword recently. As far as i know,
volatile keyword is applied on to the variable to reflect the changes done on the data of
the variable by one thread onto the other thread.
volatile keyword prevents caching of the data on the thread.
I did a small test........
I used an integer variable named count, and used volatile keyword on it.
Then made 2 different threads to increment the variable value to 10000, so the end resultant should be 20000.
But thats not the case always, with volatile keyword i am getting not getting 20000 consistently, but 18534, 15000, etc.... and sometimes 20000.
But while i used synchronized keyword, it just worked fine, why....??
Can anyone please explain me this behaviour of volatile keyword.
i am posting my code with volatile keyword and as well as the one with synchronzied keyword.
The following code below behaves inconsistently with volatile keyword on variable count
public class SynVsVol implements Runnable{
volatile int count = 0;
public void go(){
for (int i=0 ; i<10000 ; i++){
count = count + 1;
}
}
#Override
public void run() {
go();
}
public static void main(String[] args){
SynVsVol s = new SynVsVol();
Thread t1 = new Thread(s);
Thread t2 = new Thread(s);
t1.start();
t2.start();
try {
t1.join();
t2.join();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("Total Count Value: "+s.count);
}
}
The following code behaves perfectly with synchronized keyword on the method go().
public class SynVsVol implements Runnable{
int count = 0;
public synchronized void go(){
for (int i=0 ; i<10000 ; i++){
count = count + 1;
}
}
#Override
public void run() {
go();
}
public static void main(String[] args){
SynVsVol s = new SynVsVol();
Thread t1 = new Thread(s);
Thread t2 = new Thread(s);
t1.start();
t2.start();
try {
t1.join();
t2.join();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("Total Count Value: "+s.count);
}
}
count = count + 1 is not atomic. It has three steps:
read the current value of the variable
increment the value
write the new value back to the variable
These three steps are getting interwoven, resulting in different execution paths, resulting in an incorrect value. Use AtomicInteger.incrementAndGet() instead if you want to avoid the synchronized keyword.
So although the volatile keyword acts pretty much as you described it, that only applies to each seperate operation, not to all three operations collectively.
The volatile keyword is not a synchronization primitive. It merely prevents caching of the value on the thread, but it does not prevent two threads from modifying the same value and writing it back concurrently.
Let's say two threads come to the point when they need to increment the counter, which is now set to 5. Both threads see 5, make 6 out of it, and write it back into the counter. If the counter were not volatile, both threads could have assumed that they know the value is 6, and skip the next read. However, it's volatile, so they both would read 6 back, and continue incrementing. Since the threads are not going in lock-step, you may see a value different from 10000 in the output, but there's virtually no chance that you would see 20000.
The fact that a variable is volatile does not mean every operation it's involved in is atomic. For instance, this line in SynVsVol.Go:
count = count + 1;
will first have count read, then incremented, and the result will then be written back. If some other thread will execute it at the same time, the results depend on the interleaving of the commands.
Now, when you add the syncronized, SynVsVol.Go executes atomically. Namely, the increment is done as a whole by a single thread, and the other one can't modify count until it is done.
Lastly, caching of member variables that are only modified within a syncronized block is much easier. The compiler can read their value when the monitor is acquired, cache it in a register, have all changes done on that register, and eventually flush it back to the main memory when the monitor is released. This is also the case when you call wait in a synchronized block, and when some other thread notifys you: cached member variables will be synchronized, and your program will remain coherent. That's guaranteed even if the member variable is not declared as volatile:
Synchronization ensures that memory writes by a thread before or
during a synchronized block are made visible in a predictable manner
to other threads which synchronize on the same monitor.
Your code is broken because it treats the read-and-increment operation on a volatile as atomic, which it is not. The code doesn't contain a data race, but it does contain a race condition on the int.