I'm learning multithreaded counter and I'm wondering why no matter how many times I ran the code it produces the right result.
public class MainClass {
public static void main(String[] args) {
Counter counter = new Counter();
for (int i = 0; i < 3; i++) {
CounterThread thread = new CounterThread(counter);
thread.start();
}
}
}
public class CounterThread extends Thread {
private Counter counter;
public CounterThread(Counter counter) {
this.counter = counter;
}
public void run() {
for (int i = 0; i < 10; i++) {
this.counter.add();
}
this.counter.print();
}
}
public class Counter {
private int count = 0;
public void add() {
this.count = this.count + 1;
}
public void print() {
System.out.println(this.count);
}
}
And this is the result
10
20
30
Not sure if this is just a fluke or is this expected? I thought the result is going to be
10
10
10
Try increasing the loop count from 10 to 10000 and you'll likely see some differences in the output.
The most logical explanation is that with only 10 additions, a thread is too fast to finish before the next thread gets started and adds on top of the previous result.
I'm learning multithreaded counter and I'm wondering why no matter how many times I ran the code it produces the right result.
<ttdr> Check out #manouti's answer. </ttdr>
Even though you are sharing the same Counter object, which is unsynchronized, there are a couple of things that are causing your 3 threads to run (or look like they are running) serially with data synchronization. I had to work hard on my 8 proc Intel Linux box to get it to show any interleaving.
When threads start and when they finish, there are memory barriers that are crossed. According to the Java Memory Model, the guarantee is that the thread that does the thread.join() will see the results of the thread published to it but I suspect a central memory flush happens when the thread finishes. This means that if the threads run serially (and with such a small loop it's hard for them not to) they will act as if there is no concurrency because they will see each other's changes to the Counter.
Putting a Thread.sleep(100); at the front of the thread run() method causes it to not run serially. It also hopefully causes the threads to cache the Counter and not see the results published by other threads that have already finished. Still needed help though.
Starting the threads in a loop after they all have been instantiated helps concurrency.
Another thing that causes synchronization is:
System.out.println(this.count);
System.out is a Printstream which is a synchronized class. Every time a thread calls println(...) it is publishing its results to central memory. If you instead recorded the value and then displayed it later, it might show better interleaving.
I really wonder if some Java compiler inlining of the Counter class at some point is causing part of the artificial synchronization. For example, I'm really surprised that a Thread.sleep(1000) at the front and end of the thread.run() method doesn't show 10,10,10.
It should be noted that on a non-intel architecture, with different memory and/or thread models, this might be easier to reproduce.
Oh, as commentary and apropos of nothing, typically it is recommended to implement Runnable instead of extending Thread.
So the following is my tweaks to your test program.
public class CounterThread extends Thread {
private Counter counter;
int result;
...
public void run() {
try {
Thread.sleep(100);
} catch (InterruptedException e1) {
Thread.currentThread().interrupt(); // good pattern
return;
}
for (int i = 0; i < 10; i++) {
counter.add();
}
result = counter.count;
// no print here
}
}
Then your main could do something like:
Counter counter = new Counter();
List<CounterThread> counterThreads = new ArrayList<>();
for (int i = 0; i < 3; i++) {
counterThread.add(new CounterThread(counter));
}
// start in a loop after constructing them all which improves the overlap chances
for (CounterThread counterThread : counterThreads) {
counterThread.start();
}
// wait for them to finish
for (CounterThread counterThread : counterThreads) {
counterThread.join();
}
// print the results
for (CounterThread counterThread : counterThreads) {
System.out.println(counterThread.result);
}
Even with this, I never see 10,10,10 output on my box and I often see 10,20,30. Closest I get is 12,12,12.
Shows you how hard it is to properly test a threaded program. Believe me, if this code was in production and you were expecting the "free" synchronization is when it would fail you. ;-)
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.
Currently working with Threads in a production application. I want to know - If threads all have the same defined, static workload, will they complete and suspend in order? For example, if I create thread1, thread2, thread3 in a loop & have them add values up to 1,000, will they always finish in the order that they were created?
I did some testing of this theory utilizing the join() method, and the theory (on the surface) appears to be true.
public class ThreadingMain implements Runnable{
public static int total;
public static void main(String[] args) throws InterruptedException {
for(int i = 0; i < 15; i++) {
Thread t = new Thread(new ThreadingMain(), i+"");
t.start();
t.join();
}
}
#Override
public void run() {
add();
}
private static void add() {
int i = 100000;
for(int j = 0; j < i; j++) {
total += j;
}
System.out.println(Thread.currentThread().getName() + " finished");
}
}
I get the following output:
0 finished
1 finished
2 finished
...
12 finished
13 finished
14 finished
Remove the join() and this is no longer the case, as expected.
In essence, will Threads with the same workload always complete in
order? Are there any variables to this?
As a CPU has a limited amount of cores and possible executing threads at a time, the operating system needs to manage when and how long a thread gets CPU time for execution. This is done by the scheduler of the operating system.
Here you can read a bit about that: CPU Scheduling in Operating Systems
How the threads are managed by your operating system is something that you cannot influence effectively from within your code. You don't have a guarantee that, say, thread A will finish before thread B even if A was created before B.
As well, keep in mind that the threads you create are not the only ones the operating system is managing at a time. There are many more processes/programs that are not part of your code, but are involved in the scheduling.
In the tutorial of java multi-threading, it gives an exmaple of Memory Consistency Errors. But I can not reproduce it. Is there any other method to simulate Memory Consistency Errors?
The example provided in the tutorial:
Suppose a simple int field is defined and initialized:
int counter = 0;
The counter field is shared between two threads, A and B. Suppose thread A increments counter:
counter++;
Then, shortly afterwards, thread B prints out counter:
System.out.println(counter);
If the two statements had been executed in the same thread, it would be safe to assume that the value printed out would be "1". But if the two statements are executed in separate threads, the value printed out might well be "0", because there's no guarantee that thread A's change to counter will be visible to thread B — unless the programmer has established a happens-before relationship between these two statements.
I answered a question a while ago about a bug in Java 5. Why doesn't volatile in java 5+ ensure visibility from another thread?
Given this piece of code:
public class Test {
volatile static private int a;
static private int b;
public static void main(String [] args) throws Exception {
for (int i = 0; i < 100; i++) {
new Thread() {
#Override
public void run() {
int tt = b; // makes the jvm cache the value of b
while (a==0) {
}
if (b == 0) {
System.out.println("error");
}
}
}.start();
}
b = 1;
a = 1;
}
}
The volatile store of a happens after the normal store of b. So when the thread runs and sees a != 0, because of the rules defined in the JMM, we must see b == 1.
The bug in the JRE allowed the thread to make it to the error line and was subsequently resolved. This definitely would fail if you don't have a defined as volatile.
This might reproduce the problem, at least on my computer, I can reproduce it after some loops.
Suppose you have a Counter class:
class Holder {
boolean flag = false;
long modifyTime = Long.MAX_VALUE;
}
Let thread_A set flag as true, and save the time into
modifyTime.
Let another thread, let's say thread_B, read the Counter's flag. If thread_B still get false even when it is later than modifyTime, then we can say we have reproduced the problem.
Example code
class Holder {
boolean flag = false;
long modifyTime = Long.MAX_VALUE;
}
public class App {
public static void main(String[] args) {
while (!test());
}
private static boolean test() {
final Holder holder = new Holder();
new Thread(new Runnable() {
#Override
public void run() {
try {
Thread.sleep(10);
holder.flag = true;
holder.modifyTime = System.currentTimeMillis();
} catch (Exception e) {
e.printStackTrace();
}
}
}).start();
long lastCheckStartTime = 0L;
long lastCheckFailTime = 0L;
while (true) {
lastCheckStartTime = System.currentTimeMillis();
if (holder.flag) {
break;
} else {
lastCheckFailTime = System.currentTimeMillis();
System.out.println(lastCheckFailTime);
}
}
if (lastCheckFailTime > holder.modifyTime
&& lastCheckStartTime > holder.modifyTime) {
System.out.println("last check fail time " + lastCheckFailTime);
System.out.println("modify time " + holder.modifyTime);
return true;
} else {
return false;
}
}
}
Result
last check time 1565285999497
modify time 1565285999494
This means thread_B get false from Counter's flag filed at time 1565285999497, even thread_A has set it as true at time 1565285999494(3 milli seconds ealier).
The example used is too bad to demonstrate the memory consistency issue. Making it work will require brittle reasoning and complicated coding. Yet you may not be able to see the results. Multi-threading issues occur due to unlucky timing. If someone wants to increase the chances of observing issue, we need to increase chances of unlucky timing.
Following program achieves it.
public class ConsistencyIssue {
static int counter = 0;
public static void main(String[] args) throws InterruptedException {
Thread thread1 = new Thread(new Increment(), "Thread-1");
Thread thread2 = new Thread(new Increment(), "Thread-2");
thread1.start();
thread2.start();
thread1.join();
thread2.join();
System.out.println(counter);
}
private static class Increment implements Runnable{
#Override
public void run() {
for(int i = 1; i <= 10000; i++)
counter++;
}
}
}
Execution 1 output: 10963,
Execution 2 output: 14552
Final count should have been 20000, but it is less than that. Reason is count++ is multi step operation,
1. read count
2. increment count
3. store it
two threads may read say count 1 at once, increment it to 2. and write out 2. But if it was a serial execution it should have been 1++ -> 2++ -> 3.
We need a way to make all 3 steps atomic. i.e to be executed by only one thread at a time.
Solution 1: Synchronized
Surround the increment with Synchronized. Since counter is static variable you need to use class level synchronization
#Override
public void run() {
for (int i = 1; i <= 10000; i++)
synchronized (ConsistencyIssue.class) {
counter++;
}
}
Now it outputs: 20000
Solution 2: AtomicInteger
public class ConsistencyIssue {
static AtomicInteger counter = new AtomicInteger(0);
public static void main(String[] args) throws InterruptedException {
Thread thread1 = new Thread(new Increment(), "Thread-1");
Thread thread2 = new Thread(new Increment(), "Thread-2");
thread1.start();
thread2.start();
thread1.join();
thread2.join();
System.out.println(counter.get());
}
private static class Increment implements Runnable {
#Override
public void run() {
for (int i = 1; i <= 10000; i++)
counter.incrementAndGet();
}
}
}
We can do with semaphores, explicit locking too. but for this simple code AtomicInteger is enough
Sometimes when I try to reproduce some real concurrency problems, I use the debugger.
Make a breakpoint on the print and a breakpoint on the increment and run the whole thing.
Releasing the breakpoints in different sequences gives different results.
Maybe to simple but it worked for me.
Please have another look at how the example is introduced in your source.
The key to avoiding memory consistency errors is understanding the happens-before relationship. This relationship is simply a guarantee that memory writes by one specific statement are visible to another specific statement. To see this, consider the following example.
This example illustrates the fact that multi-threading is not deterministic, in the sense that you get no guarantee about the order in which operations of different threads will be executed, which might result in different observations across several runs. But it does not illustrate a memory consistency error!
To understand what a memory consistency error is, you need to first get an insight about memory consistency. The simplest model of memory consistency has been introduced by Lamport in 1979. Here is the original definition.
The result of any execution is the same as if the operations of all the processes were executed in some sequential order and the operations of each individual process appear in this sequence in the order specified by its program
Now, consider this example multi-threaded program, please have a look at this image from a more recent research paper about sequential consistency. It illustrates what a real memory consistency error might look like.
To finally answer your question, please note the following points:
A memory consistency error always depends on the underlying memory model (A particular programming languages may allow more behaviours for optimization purposes). What's the best memory model is still an open research question.
The example given above gives an example of sequential consistency violation, but there is no guarantee that you can observe it with your favorite programming language, for two reasons: it depends on the programming language exact memory model, and due to undeterminism, you have no way to force a particular incorrect execution.
Memory models are a wide topic. To get more information, you can for example have a look at Torsten Hoefler and Markus Püschel course at ETH Zürich, from which I understood most of these concepts.
Sources
Leslie Lamport. How to Make a Multiprocessor Computer That Correctly Executes Multiprocessor Programs, 1979
Wei-Yu Chen, Arvind Krishnamurthy, Katherine Yelick, Polynomial-Time Algorithms for Enforcing Sequential Consistency in SPMD Programs with Arrays, 2003
Design of Parallel and High-Performance Computing course, ETH Zürich
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'm having a-bit of trouble with threads in java. Basically Im creating an array of threads and starting them. the point of the program is to simulate a race, total the time for each competitor ( i.e. each thread ) and pick the winner.
The competitor moves one space, waits ( i.e. thread sleeps for a random period of time between 5 and 6 seconds ) and then continues. The threads don't complete in the order that they started as expected.
Now for the problem. I can get the total time it takes for a thread to complete; what I want is to store all the times from the threads into a single array and be able to calculate the fastest time.
To do this should I place the array in the main.class file? Would I be right in assuming so because if it was placed in the Thread class it wouldn't work. Or should I create a third class?
I'm alittle confused :/
It's fine to declare it in the method where you invoke the threads, with a few notes:
each thread should know its index in the array. Perhaps you should pass this in constructor
then you have three options for filling the array
the array should be final, so that it can be used within anonymous classes
the array can be passed to each thread
the threads should notify a listener when they're done, which in turn will increment an array.
consider using Java 1.5 Executors framework for submitting Runnables, rather than working directly with threads.
EDIT: The solution below assumes you need the times only after all competitors have finished the race.
You can use a structure that looks like below, (inside your main class). Typically you want to add a lot of you own stuff; this is the main outline.
Note that concurrency is not an issue at all here because you get the value from the MyRunnable instance once its thread has finished running.
Note that using a separate thread for each competitor is probably not really necessary with a modified approach, but that would be a different issue.
public static void main(String[] args) {
MyRunnable[] runnables = new MyRunnable[NUM_THREADS];
Thread[] threads = new Thread[NUM_THREADS];
for (int i = 0; i < NUM_THREADS; i++) {
runnables[i] = new MyRunnable();
threads[i] = new Thread(runnables[i]);
}
// start threads
for (Thread thread : threads) {
thread.start();
}
// wait for threads
for (Thread thread : threads) {
try {
thread.join();
} catch (InterruptedException e) {
// ignored
}
}
// get the times you calculated for each thread
for (int i = 0; i < NUM_THREADS; i++) {
int timeSpent = runnables[i].getTimeSpent();
// do something with the time spent
}
}
static class MyRunnable implements Runnable {
private int timeSpent;
public MyRunnable(...) {
// initialize
}
public void run() {
// whatever the thread should do
// finally set the time
timeSpent = ...;
}
public int getTimeSpent() {
return timeSpent;
}
}