Here is my code,
class Shared {
private static int index = 0;
public synchronized void printThread() {
try {
while(true) {
Thread.sleep(1000);
System.out.println(Thread.currentThread().getName() + ": " + index++);
notifyAll();
// notify();
wait();
}
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
class Example13 implements Runnable {
private Shared shared = new Shared();
#Override
public void run() {
shared.printThread();
}
}
public class tetest {
public static void main(String[] args) {
Example13 r = new Example13();
Thread t1 = new Thread(r, "Thread 1");
Thread t2 = new Thread(r, "Thread 2");
Thread t3 = new Thread(r, "Thread 3");
Thread t4 = new Thread(r, "Thread 4");
Thread t5 = new Thread(r, "Thread 5");
t1.start();
t2.start();
t3.start();
t4.start();
t5.start();
}
}
and the result is here
Thread 1: 0
Thread 5: 1
Thread 4: 2
Thread 3: 3
Thread 2: 4
Thread 3: 5
Thread 2: 6
Thread 3: 7
Thread 2: 8
Thread 3: 9
the question is, why only two of the threads are working? I'm so confused, I thought notify() is randomly wake up one of the waiting threads, but it's not.
is this starvation? then, why the starvation is caused? I tried notify() and notifyAll(), but got same results for both.
Can any one help my toasted brain?
This isn't 'starvation'. Your 5 threads are all doing nothing. They all want to 'wake up' - notify() will wake up an arbitrary one. It is neither unreliably random: The JMM does not ascribe an order to this, so one of them will wake up, you can't rely on it being random (do not use this to generate random numbers), nor can you rely on specific ordering behaviour.
It's not starvation (it's not: Oh no! Thread 2 and 3 are doing all the work and 4 and 5 are just hanging out doing nothing! That's bad - the system could be more efficient!) - because it doesn't matter which thread 'does the work'. A CPU core is a CPU core, it cares not which thread it ends up running.
Starvation is a different principle. Imagine instead of Thread.sleep (which means the threads aren't waiting for anything specific, other than some time to elapse), instead the threads want to print the result of some expensiv-ish math operation. If you just let 2 threads each say 'Hello!', then the impl of System.out says that it would be acceptable for the JVM to produce:
HelHellloo!!
So to prevent this, you use locks to create a 'talking stick' of sorts: A thread only gets to print if it has the talking stick. Each of 5 threads will all perform, in a loop, the following operation:
Do an expensive math operation.
Acquire the talking stick.
Print the result of the operation.
Release the talking stick.
Loop back to the top.
Now imagine that, despite the fact that the math operation is quite expensive, for whatever reason you have an excruciatingly slow terminal, and the 'print the result of the operation' job takes a really long time to finish.
Now you can run into starvation. Imagine this scenario:
Threads 1-5 all do their expensive math simultaneously.
Arbitrarily, thread 4 ends up nabbing the talking stick.
The other 4 threads soon too want the talking stick but they have to wait; t4 has it. They do nothing now. Twiddling their thumbs (they could be calculating, but they are not!)
after the excruciatingly long time, 4 is done and releases the stick. 1, 2, 3, and 5 dogpile on like its the All-Blacks and 2 so happens to win the scrum and crawls out of the pile with the stick. 1, 3, and 5 gnash their teeth and go back yet again to waiting for the stick, still not doing any work. Whilst 2 is busy spending the really long time printing results, 4 goes back to the top of the loop and calculates another result. It ends up doing this faster than 2 manages to print, so 4 ends up wanting the talking stick again before 2 is done.
2 is finally done and 1, 3, 4, and 5 all scrum into a pile again. 4 happens to get the stick - java makes absolutely no guarantees about fairness, any one of them can get it, there is also no guarantee of randomness or lack thereof. A JVM is not broken if 4 is destined to win this fight.
Repeat ad nauseam. 2 and 4 keep trading the stick back and forth. 1, 3, and 5 never get to talk.
The above is, as per the JMM, valid - a JVM is not broken if it behaves like this (it would be a tad weird). Any bugs filed about this behaviour would be denied: The lock isn't so-called "fair". Java has fair locks if you want them - in the java.util.concurrent package. Fair locks incur some slight extra bookkeeping cost, the assumption made by the synchronized and wait/notify system is that you don't want to pay this extra cost.
A better solution to the above scenario is possibly to make a 6th thread that JUST prints, with 5 threads JUST filling a buffer, at least then the 'print' part is left to a single thread and that might be faster. But mostly, the bottleneck in this getup is simply that printing - the code has zero benefit from being multicore (just having ONE thread do ONE math calculation, print it, do another, and so forth would be better. Or possibly 2 threads: Whilst printing, the other thread calculates a number, but there's no point in having more than one; even a single thread can calculate faster than results can be printed). Thus in some ways this is just what the situation honestly requires: This hypothetical scenario still prints as fast as it can. IF you need the printing to be 'fair' (and who says that? It's not intrinsic to the problem description that fairness is a requirement. Maybe all the various calculations are equally useful so it doesn't matter that one thread gets to print more than others; let's say its bitcoin miners, generating a random number and checking if that results in a hash with the requisite 7 zeroes at the end or whatever bitcoin is up to now - who cares that one thread gets more time than another? A 'fair' system is no more likely to successfully mine a block).
Thus, 'fairness' is something you need to explicitly determine you actually need. If you do, AND starvation is an issue, use a fair lock. new ReentrantLock(true) is all you need (that boolean parameter is the fair parameter - true means you want fairness).
Related
An optimum of threads in a pool is something that is case specific, though there is a rule of thumb which says #threads = #CPU +1.
However, how does this work with threads spanning other threads and waiting (i.e. blocked until thread.join() is successful) for these 'subthreads'?
Assume that I have code that requires the execution of list of tasks (2), which has subtasks(2), which has subsubtasks(3) and so on. The total number of tasks is 2*2*3 = 12, though 18 threads will be created (because a threads will 'spawn' more subtasks (threads), where the thread spawning more threads will be blocked untill all is over. See below for pseudo code.
I am assuming that for a CPU with N cores there is a rule of thumb that everything can be parallelized if the highest number of active threads (12) is #CPU + 1. Is this correct?
PseudoCode
outputOfTask = []
for subtask in SubTaskList
outputOfTask --> append(subtask.doCompute())
// wait untill all output is finished.
in subtask.java:
Each subtask, for example, implements the same interface, but can be different.
outputOfSubtask = []
for task in subsubTaskList
// do some magic depending on the type of subtask
outputOfSubtask -> append( task.doCompute())
return outputOfSubtask
in subsubtask.java:
outputOfSubsubtask = []
for task in subsubsubtask
// do some magic depending on the type of subsubtask
outputOfSubsubtask -> append( task.doCompute())
return outputOfSubsubtask
EDIT:
Dummy code Java code. I used this in my original question to check how many threads were active, but I assume that the pseudocode is more clear. Please note: I used the Eclipse Collection, this introduces the asParallel function which allows for a shorter notation of the code.
#Test
public void testasParallelthreads() {
// // ExecutorService executor = Executors.newWorkStealingPool();
ExecutorService executor = Executors.newCachedThreadPool();
MutableList<Double> myMainTask = Lists.mutable.with(1.0, 2.0);
MutableList<Double> mySubTask = Lists.mutable.with(1.0, 2.0);
MutableList<Double> mySubSubTask = Lists.mutable.with(1.0, 2.0);
MutableList<Double> mySubSubSubTask = Lists.mutable.with(1.0, 2.0, 2.0);
MutableList<Double> a = myMainTask.asParallel(executor, 1)
.flatCollect(task -> mySubTask.asParallel(executor,1)
.flatCollect(subTask -> mySubSubTask.asParallel(executor, 1)
.flatCollect(subsubTask -> mySubSubSubTask.asParallel(executor, 1)
.flatCollect(subsubTask -> dummyFunction(task, subTask, subsubTask, subsubTask,executor))
.toList()).toList()).toList()).toList();
System.out.println("pool size: " + ((ThreadPoolExecutor) executor).getPoolSize());
executor.shutdownNow();
}
private MutableList<Double> dummyFunction(double a, double b, double c, double d, ExecutorService ex) {
System.out.println("ThreadId: " + Thread.currentThread().getId());
System.out.println("Active threads size: " + ((ThreadPoolExecutor) ex).getActiveCount());
return Lists.mutable.with(a,b,c,d);
}
I am assuming that for a CPU with N cores there is a rule of thumb that everything can be parallelized if the highest number of active threads (12) is #CPU + 1. Is this correct?
This topic is extremely hard to generalize about. Even with the actual code, the performance of your application is going to be very difficult to determine. Even if you could come up an estimation, the actual performance may vary wildly between runs – especially considering that the threads are interacting with each other. The only time we can take the #CPU + 1 number is if the jobs that are submitted into the thread-pool are independent and completely CPU bound.
I'd recommend trying a number of different thread-pool size values under simulated load to find the optimal values for your application. Examining the overall throughput numbers or system load stats should give you the feedback you need.
However, how does this work with threads spanning other threads and waiting (i.e. blocked until thread.join() is successful) for these 'subthreads'?
Threads will block, and it is up to the os/jvm to schedule another one if possible. If you have a single thread pool executor and call join from one of your tasks, the other task won't even get started. With executors that use more threads, then the blocking task will block a single thread and the os/jvm is free to scheduled other threads.
These blocked threads should not consume CPU time, because they are blocked. So I am assuming that for a CPU with N cores there is a rule of thumb that everything can be parallelized if the highest number of active threads (24) is #CPU + 1. Is this correct?
Active threads can be blocking. I think you're mixing terms here, #CPU, the number of cores, and the number of virtual cores. If you have N physical cores, then you can run N cpu bound tasks in parallel. When you have other types of blocking or very short lived tasks, then you can have more parallel tasks.
I am trying to learn concurrency in Java, but whatever I do, 2 threads run in serial, not parallel, so I am not able to replicate common concurrency issues explained in tutorials (like thread interference and memory consistency errors). Sample code:
public class Synchronization {
static int v;
public static void main(String[] args) {
Runnable r0 = () -> {
for (int i = 0; i < 10; i++) {
Synchronization.v++;
System.out.println(v);
}
};
Runnable r1 = () -> {
for (int i = 0; i < 10; i++) {
Synchronization.v--;
System.out.println(v);
}
};
Thread t0 = new Thread(r0);
Thread t1 = new Thread(r1);
t0.start();
t1.start();
}
}
This always give me a result starting from 1 and ending with 0 (whatever the loop length is). For example, the code above gives me every time:
1
2
3
4
5
6
7
8
9
10
9
8
7
6
5
4
3
2
1
0
Sometimes, the second thread starts first and the results are the same but negative, so it is still running in serial.
Tried in both Intellij and Eclipse with identical results. CPU has 2 cores if it matters.
UPDATE: it finally became reproducible with huge loops (starting from 1_000_000), though still not every time and just with small amount of final discrepancy. Also seems like making operations in loops "heavier", like printing thread name makes it more reproducible as well. Manually adding sleep to thread also works, but it makes experiment less cleaner, so to say. The reason doesn't seems to be that first loop finishes before the second starts, because I see both loops printing to console while continuing operating and still giving me 0 at the end. The reasons seems more like a thread race for same variable. I will dig deeper into that, thanks.
Seems like first started thread just never give a chance to second in Thread Race to take a variable/second one just never have a time to even start (couldn't say for sure), so the second almost* always will be waiting until first loop will be finished.
Some heavy operation will mix the result:
TimeUnit.MILLISECONDS.sleep(100);
*it is not always true, but you are was lucky in your tests
Starting a thread is heavyweight operation, meaning that it will take some time to perform. Due that fact, by the time you start second thread, first is finished.
The reasoning why sometimes it is in "revert order" is due how thread scheduler works. By the specs there are not guarantees about thread execution order - having that in mind, we know that it is possible for second thread to run first (and finish)
Increase iteration count to something meaningful like 10000 and see what will happen then.
This is called lucky timing as per Brian Goetz (Author of Java Concurrency In Practice). Since there is no synchronization to the static variable v it is clear that this class is not thread-safe.
I've stumbled upon a modified version of the Bakerys Algorithm (an uncomplete one of course with flaws)
I've been asked in class if the following algorithm is can have a starvation issue:
while(true){
number[me] = max(number[0],...,number[n]) + 1
for (other from 0 to n) {
while(number[other] != 0 && number[other] < number[me]) {
// Wait
}
}
/*CS*/
number[me] = 0
}
I understand that a deadlock is possible however, I'm asking is this algorithm starvation-free ?
I think that it is, because I can guarantee that once thread A has chosen a number, other threads will always have a bigger number than thread A and therefor he will eventually be allowed to enter the CS
My friend thinks that the algorithm is not starvation free, since a thread can be stuck in the process of taking a number (calculating the max) and possibly get its CPU time taken from him. Meanwhile other threads will start & finish and perhaps start again (since the while true) while supposedly thread A is being starved.
My question can be simplified to this:
Does the choosing array in the original Bakerys Algorithm solve starvation ?
Starvation-freedom can be defined as: Any process trying to enter critical section, will eventually enter critical section.
The line that calculates max is not part of the critical section, so it will eventually receive cpu time to make that assignment.
When a process A receives its id, then it will wait for all the other process that has an id lower than the one it has (lower id means that has more priority). Sometime that processes will leave the critical section and will get a new id. This id will be greater than the one it has and in that moment process A will enter in the critical section.
Finally, the algorithm is starvation-free.
I am slightly surprised by what I get if I compile and run the following (horrible non-synchronized) Java SE program.
public class ThreadRace {
// this is the main class.
public static void main(String[] args) {
TestRunnable tr=new TestRunnable(); // tr is a Runnable.
Thread one=new Thread(tr,"thread_one");
Thread two=new Thread(tr,"thread_two");
one.start();
two.start(); // starting two threads both with associated object tr.
}
}
class TestRunnable implements Runnable {
int counter=0; // Both threads can see this counter.
public void run() {
for(int x=0;x<1000;x++) {
counter++;
}
// We can't get here until we've added one to counter 1000 times.
// Can we??
System.out.println("This is thread "+
Thread.currentThread().getName()+" and the counter is "+counter);
}
}
If I run "java ThreadRace" at the command line, then here is my interpretation
of what happens. Two new threads are created and started. The threads have
the same Runnable object instance tr, and so they see the same tr.counter .
Both new threads add one to this counter 1000 times, and then print the value
of the counter.
If I run this lots and lots of times, then usually I get output of the form
This is thread thread_one and the counter is 1000
This is thread thread_two and the counter is 2000
and occasionally I get output of the form
This is thread thread_one and the counter is 1204
This is thread thread_two and the counter is 2000
Note that what happened in this latter case was that thread_one finished
adding one to the counter 1000 times, but thread_two had started adding
one already, before thread_one printed out the value of the counter.
In particular, this output is still comprehensible to me.
However, very occasionally I get something like
This is thread thread_one and the counter is 1723
This is thread thread_two and the counter is 1723
As far as I can see, this "cannot happen". The only way the System.out.println() line
can be reached in either thread, is if the thread has finished counting to 1000.
So I am not bothered if one of the threads reports the counter as being some
random number between 1000 and 2000, but I cannot see how both threads can
get as far as their System.out.println() line (implying both for loops have finished,
surely?) and counter not be 2000 by the time the second statement is printed.
Is what is happening that both threads somehow attempt to do counter++ at exactly
the same time, and one overwrites the other? That is, a thread can even be
interrupted even in the middle of executing a single statement?
The "++" operator is not atomic -- it doesn't happen in one uninterruptible cycle. Think of it like this:
1. Fetch the old value
2. Add one to it
3. Store the new value back
So imagine that you get this sequence:
Thread A: Step 1
Thread B: Step 1
Thread A: Step 2
Thread B: Step 2
Thread A: Step 3
Thread B: Step 3
Both threads think they've incremented the variable, but its value has only increased by one! The second "store back" operation effectively cancels out the result of the first.
Now, truth is, when you add in multiple levels of cache, far weirder things can actually happen; but this is an easy explanation to understand. You can fix these kinds of issues by synchronizing access to the variable: either the whole run() method, or the inside of the loop using a synchronized block. As Jon suggests, you could also use some of the fancier tools in java.util.concurrent.atomic.
It absolutely can happen. counter++ isn't an atomic operation. Consider it as:
int tmp = counter;
tmp++;
counter = tmp;
Now imagine two threads executing that code at the same about time:
Both read the counter
Both increment their local copy (0 to 1)
Both write 1 into counter
This sort of thing is precisely why java.util.concurrent.atomic exists. Change your code to:
class TestRunnable implements Runnable {
private final AtomicInteger counter = new AtomicInteger();
public void run() {
for(int x=0;x<1000;x++) {
counter.incrementAndGet();
}
System.out.println("This is thread "+
Thread.currentThread().getName()+" and the counter is " + counter.get());
}
}
That code is safe.
I just have implemented a threaded version of the merge sort. ThreadedMerge.java: http://pastebin.com/5ZEvU6BV
Since merge sort is a divide and conquer algorithm I create a thread for every half of the array. But the number of avialable threads in Java-VM is limited so I check that before creating threads:
if(num <= nrOfProcessors){
num += 2;
//create more threads
}else{
//continue without threading
}
However the threaded sorting takes about ~ 6000 ms while the non-threaded version is much faster with just ~ 2500 ms.
Non-Threaded: http://pastebin.com/7FdhZ4Fw
Why is the threaded version slower and how do I solve that problem?
Update: I use atomic integer now for thread counting and declared a static field for Runtime.getRuntime().availableProcessors(). The sorting takes about ~ 1400 ms now.
However creating just one thread in the mergeSort method and let the current thread do the rest has no sigificant performance increase. Why?
Besides when after I call join on a thread and after that decrement the number of used threads with
num.set(num.intValue() - 1);
the sorting takes about ~ 200 ms longer. Here is the update of my algorithm http://pastebin.com/NTZq5zQp Why does this line of code make it even worse?
first off your accesses to num is not threadsafe (check http://download.oracle.com/javase/6/docs/api/java/util/concurrent/atomic/AtomicInteger.html )
you create an equal amount of processes to cores but you block half of them with the join call
num += 1;
ThreadedMerge tm1 = new ThreadedMerge(array, startIndex, startIndex + halfLength);
tm1.start();
sortedRightPart = mergeSort(array, startIndex + halfLength, endIndex);
try{
tm1.join();
num-=1
sortedLeftPart = tm1.list;
}catch(InterruptedException e){
}
this doesn't block the calling thread but uses it to sort the right part and let the created thread do the other part when that one returns the space it takes up can be used by another thread
Hhm, you should not create a thread for every single step (they are expensive and there are lightweight alternatives.)
Ideally, you should only create 4 threads if there are 4 CPU´s.
So let´s say you have 4 CPU´s, then you create one thread at the first level (now you have 2) and at the second level you also create a new thread. This gives you 4.
The reason why you only create one and not two is that you can use the thread you are currently running like:
Thread t = new Thread(...);
t.start();
// Do half of the job here
t.join(); // Wait for the other half to complete.
If you have, let´s say, 5 CPU´s (not in the power of two) then just create 8 threads.
One simple way to do this in practice, is to create the un-threaded version you already made when you reach the appropriate level. In this way you avoid to clutter the merge method when if-sentences etc.
The call to Runtime.availableProcessors() appears to be taking up a fair amount of extra time. You only need to call it once, so just move it outside of the method and define it as a static, e.g.:
static int nrOfProcessors = Runtime.getRuntime().availableProcessors();