Does an object always see its latest internal state irrespective of thread?





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11















Let's say I have a runnable with a simple integer count variable which is incremented every time runnable runs. One instance of this object is submitted to run periodically in a scheduled executor service.



class Counter implements Runnable {
private int count = 0;

@Override
public void run() {
count++;
}
}

Counter counter = new Counter();
ScheduledExecutorService executorService = Executors.newScheduledThreadPool(5);
executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


Here, the object is accessing its own internal state inside of different threads (reading and incrementing). Is this code thread-safe or is it possible that we lose updates to the count variable when it's scheduled in a different thread?










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  • 1





    Nope; most definitely not.

    – Boris the Spider
    yesterday






  • 3





    Just to be clear: When the answers below say "...no happens-before..." what they're saying is, Suppose count==N. Then along comes worker thread A, which sets count = N+1. Then one whole second later, worker thread B is chosen to call the run() method, and worker thread B looks at count. It is possible at that point for worker thread B to still see count == N.

    – Solomon Slow
    yesterday




















11















Let's say I have a runnable with a simple integer count variable which is incremented every time runnable runs. One instance of this object is submitted to run periodically in a scheduled executor service.



class Counter implements Runnable {
private int count = 0;

@Override
public void run() {
count++;
}
}

Counter counter = new Counter();
ScheduledExecutorService executorService = Executors.newScheduledThreadPool(5);
executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


Here, the object is accessing its own internal state inside of different threads (reading and incrementing). Is this code thread-safe or is it possible that we lose updates to the count variable when it's scheduled in a different thread?










share|improve this question




















  • 1





    Nope; most definitely not.

    – Boris the Spider
    yesterday






  • 3





    Just to be clear: When the answers below say "...no happens-before..." what they're saying is, Suppose count==N. Then along comes worker thread A, which sets count = N+1. Then one whole second later, worker thread B is chosen to call the run() method, and worker thread B looks at count. It is possible at that point for worker thread B to still see count == N.

    – Solomon Slow
    yesterday
















11












11








11


2






Let's say I have a runnable with a simple integer count variable which is incremented every time runnable runs. One instance of this object is submitted to run periodically in a scheduled executor service.



class Counter implements Runnable {
private int count = 0;

@Override
public void run() {
count++;
}
}

Counter counter = new Counter();
ScheduledExecutorService executorService = Executors.newScheduledThreadPool(5);
executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


Here, the object is accessing its own internal state inside of different threads (reading and incrementing). Is this code thread-safe or is it possible that we lose updates to the count variable when it's scheduled in a different thread?










share|improve this question
















Let's say I have a runnable with a simple integer count variable which is incremented every time runnable runs. One instance of this object is submitted to run periodically in a scheduled executor service.



class Counter implements Runnable {
private int count = 0;

@Override
public void run() {
count++;
}
}

Counter counter = new Counter();
ScheduledExecutorService executorService = Executors.newScheduledThreadPool(5);
executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


Here, the object is accessing its own internal state inside of different threads (reading and incrementing). Is this code thread-safe or is it possible that we lose updates to the count variable when it's scheduled in a different thread?







java multithreading concurrency






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share|improve this question













share|improve this question




share|improve this question








edited yesterday









Peter Mortensen

13.9k1987113




13.9k1987113










asked yesterday









RandomQuestionRandomQuestion

3,067144580




3,067144580








  • 1





    Nope; most definitely not.

    – Boris the Spider
    yesterday






  • 3





    Just to be clear: When the answers below say "...no happens-before..." what they're saying is, Suppose count==N. Then along comes worker thread A, which sets count = N+1. Then one whole second later, worker thread B is chosen to call the run() method, and worker thread B looks at count. It is possible at that point for worker thread B to still see count == N.

    – Solomon Slow
    yesterday
















  • 1





    Nope; most definitely not.

    – Boris the Spider
    yesterday






  • 3





    Just to be clear: When the answers below say "...no happens-before..." what they're saying is, Suppose count==N. Then along comes worker thread A, which sets count = N+1. Then one whole second later, worker thread B is chosen to call the run() method, and worker thread B looks at count. It is possible at that point for worker thread B to still see count == N.

    – Solomon Slow
    yesterday










1




1





Nope; most definitely not.

– Boris the Spider
yesterday





Nope; most definitely not.

– Boris the Spider
yesterday




3




3





Just to be clear: When the answers below say "...no happens-before..." what they're saying is, Suppose count==N. Then along comes worker thread A, which sets count = N+1. Then one whole second later, worker thread B is chosen to call the run() method, and worker thread B looks at count. It is possible at that point for worker thread B to still see count == N.

– Solomon Slow
yesterday







Just to be clear: When the answers below say "...no happens-before..." what they're saying is, Suppose count==N. Then along comes worker thread A, which sets count = N+1. Then one whole second later, worker thread B is chosen to call the run() method, and worker thread B looks at count. It is possible at that point for worker thread B to still see count == N.

– Solomon Slow
yesterday














3 Answers
3






active

oldest

votes


















8














Does an object always see its latest internal state irrespective of thread?



Just to be clear for the purposes of this question and its answers, an object doesn't do anything; it's just memory. Threads are the executing entity. It's misleading to say does an object see whatever. It's the thread that's doing the seeing/reading of object state.



This isn't specified in the javadoc, but



Executors.newScheduledThreadPool(5);


returns a ScheduledThreadPoolExecutor.



Your code is using



executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


The javadoc for ScheduledThreadPoolExecutor#scheduledWithFixedDelay states




Submits a periodic action that becomes enabled first after the given
initial delay, and subsequently with the given delay between the
termination of one execution and the commencement of the next.




The class javadoc further clarifies




Successive executions of a periodic task scheduled via
scheduleAtFixedRate or scheduleWithFixedDelay do not overlap.
While different executions may be performed by different threads, the
effects of prior executions happen-before those of subsequent ones
.




As such, each execution of Counter#run is guaranteed to see the value of count after it's been incremented by the previous execution. For example, the third execution will read a count value of 2 before it performs its increment.



You don't need volatile or any other additional synchronization mechanism for this specific use case.






share|improve this answer


























  • Thank you for pointing out this happens-before guarantee about ScheduledThreadPoolExecutor.

    – RandomQuestion
    yesterday



















6














No, this code is not thread-safe because there isn't any happens-before relation between increments made in different threads started with ScheduledExecutorService.



To fix it, you need to either mark the variable as volatile or switch to AtomicInteger or AtomicLong.



UPDATE:



As @BoristheSpider mentioned, in general in case of increment/decrement making a variable volatile is not enough since increment/decrement is not atomic itself and calling it from several threads concurrently will cause race conditions and missed updates. However, in this particular case scheduleWithFixedDelay() guarantees (according to Javadoc) that there will be overlapping executions of scheduled task, so volatile will also work in this particular case even with increment.






share|improve this answer


























  • @BoristheSpider taking into account how scheduleWithFixedDelay work there will be no overlapped calls to counter++ in that particular scenario. So volatile should be OK.

    – Ivan
    yesterday













  • There is a happens-before relation introduced between subsequent execution of a task scheduled with scheduleWithFixedDelay.

    – Sotirios Delimanolis
    yesterday











  • How come volatile is not enough? The semantics of volatile should hold in every case and it would solve this problem. In the past, the writing of doubles and longs could have been more difficult to do atomically since those occupied more than one 32-bit register. However, as far as I can tell, that should not affect the semantics of volatile. Or am I missing something?

    – Edwin Dalorzo
    yesterday






  • 1





    @EdwinDalorzo in this particular case volatile is enough. But in general case not because increment is effectively translated into 3 operations: read, add, write and is prone to race condition in multithreaded environment

    – Ivan
    yesterday











  • To reiterate, the code they have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday



















2














No, this code is not thread-safe since there isn't any happens before relation between different threads accessing count.






share|improve this answer





















  • 1





    Even then it's not thread safe, because ++ isn't atomic.

    – Andy Turner
    yesterday






  • 1





    @michid please correct me if I am wrong but shouldn't counter++ be synchronized too as increment operation is not atomic.

    – Yug Singh
    yesterday






  • 1





    Does counter++ need to be synchronized even if it's guaranteed that only one thread will be running this runnable at a time (like in the code in the question)?

    – RandomQuestion
    yesterday













  • @RandomQuestion yes. Because visibility.

    – Boris the Spider
    yesterday








  • 1





    @RandomQuestion To reiterate, the code you have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









8














Does an object always see its latest internal state irrespective of thread?



Just to be clear for the purposes of this question and its answers, an object doesn't do anything; it's just memory. Threads are the executing entity. It's misleading to say does an object see whatever. It's the thread that's doing the seeing/reading of object state.



This isn't specified in the javadoc, but



Executors.newScheduledThreadPool(5);


returns a ScheduledThreadPoolExecutor.



Your code is using



executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


The javadoc for ScheduledThreadPoolExecutor#scheduledWithFixedDelay states




Submits a periodic action that becomes enabled first after the given
initial delay, and subsequently with the given delay between the
termination of one execution and the commencement of the next.




The class javadoc further clarifies




Successive executions of a periodic task scheduled via
scheduleAtFixedRate or scheduleWithFixedDelay do not overlap.
While different executions may be performed by different threads, the
effects of prior executions happen-before those of subsequent ones
.




As such, each execution of Counter#run is guaranteed to see the value of count after it's been incremented by the previous execution. For example, the third execution will read a count value of 2 before it performs its increment.



You don't need volatile or any other additional synchronization mechanism for this specific use case.






share|improve this answer


























  • Thank you for pointing out this happens-before guarantee about ScheduledThreadPoolExecutor.

    – RandomQuestion
    yesterday
















8














Does an object always see its latest internal state irrespective of thread?



Just to be clear for the purposes of this question and its answers, an object doesn't do anything; it's just memory. Threads are the executing entity. It's misleading to say does an object see whatever. It's the thread that's doing the seeing/reading of object state.



This isn't specified in the javadoc, but



Executors.newScheduledThreadPool(5);


returns a ScheduledThreadPoolExecutor.



Your code is using



executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


The javadoc for ScheduledThreadPoolExecutor#scheduledWithFixedDelay states




Submits a periodic action that becomes enabled first after the given
initial delay, and subsequently with the given delay between the
termination of one execution and the commencement of the next.




The class javadoc further clarifies




Successive executions of a periodic task scheduled via
scheduleAtFixedRate or scheduleWithFixedDelay do not overlap.
While different executions may be performed by different threads, the
effects of prior executions happen-before those of subsequent ones
.




As such, each execution of Counter#run is guaranteed to see the value of count after it's been incremented by the previous execution. For example, the third execution will read a count value of 2 before it performs its increment.



You don't need volatile or any other additional synchronization mechanism for this specific use case.






share|improve this answer


























  • Thank you for pointing out this happens-before guarantee about ScheduledThreadPoolExecutor.

    – RandomQuestion
    yesterday














8












8








8







Does an object always see its latest internal state irrespective of thread?



Just to be clear for the purposes of this question and its answers, an object doesn't do anything; it's just memory. Threads are the executing entity. It's misleading to say does an object see whatever. It's the thread that's doing the seeing/reading of object state.



This isn't specified in the javadoc, but



Executors.newScheduledThreadPool(5);


returns a ScheduledThreadPoolExecutor.



Your code is using



executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


The javadoc for ScheduledThreadPoolExecutor#scheduledWithFixedDelay states




Submits a periodic action that becomes enabled first after the given
initial delay, and subsequently with the given delay between the
termination of one execution and the commencement of the next.




The class javadoc further clarifies




Successive executions of a periodic task scheduled via
scheduleAtFixedRate or scheduleWithFixedDelay do not overlap.
While different executions may be performed by different threads, the
effects of prior executions happen-before those of subsequent ones
.




As such, each execution of Counter#run is guaranteed to see the value of count after it's been incremented by the previous execution. For example, the third execution will read a count value of 2 before it performs its increment.



You don't need volatile or any other additional synchronization mechanism for this specific use case.






share|improve this answer















Does an object always see its latest internal state irrespective of thread?



Just to be clear for the purposes of this question and its answers, an object doesn't do anything; it's just memory. Threads are the executing entity. It's misleading to say does an object see whatever. It's the thread that's doing the seeing/reading of object state.



This isn't specified in the javadoc, but



Executors.newScheduledThreadPool(5);


returns a ScheduledThreadPoolExecutor.



Your code is using



executorService.scheduleWithFixedDelay(counter, 1, 1, TimeUnit.SECONDS);


The javadoc for ScheduledThreadPoolExecutor#scheduledWithFixedDelay states




Submits a periodic action that becomes enabled first after the given
initial delay, and subsequently with the given delay between the
termination of one execution and the commencement of the next.




The class javadoc further clarifies




Successive executions of a periodic task scheduled via
scheduleAtFixedRate or scheduleWithFixedDelay do not overlap.
While different executions may be performed by different threads, the
effects of prior executions happen-before those of subsequent ones
.




As such, each execution of Counter#run is guaranteed to see the value of count after it's been incremented by the previous execution. For example, the third execution will read a count value of 2 before it performs its increment.



You don't need volatile or any other additional synchronization mechanism for this specific use case.







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday









Peter Mortensen

13.9k1987113




13.9k1987113










answered yesterday









Sotirios DelimanolisSotirios Delimanolis

213k41500590




213k41500590













  • Thank you for pointing out this happens-before guarantee about ScheduledThreadPoolExecutor.

    – RandomQuestion
    yesterday



















  • Thank you for pointing out this happens-before guarantee about ScheduledThreadPoolExecutor.

    – RandomQuestion
    yesterday

















Thank you for pointing out this happens-before guarantee about ScheduledThreadPoolExecutor.

– RandomQuestion
yesterday





Thank you for pointing out this happens-before guarantee about ScheduledThreadPoolExecutor.

– RandomQuestion
yesterday













6














No, this code is not thread-safe because there isn't any happens-before relation between increments made in different threads started with ScheduledExecutorService.



To fix it, you need to either mark the variable as volatile or switch to AtomicInteger or AtomicLong.



UPDATE:



As @BoristheSpider mentioned, in general in case of increment/decrement making a variable volatile is not enough since increment/decrement is not atomic itself and calling it from several threads concurrently will cause race conditions and missed updates. However, in this particular case scheduleWithFixedDelay() guarantees (according to Javadoc) that there will be overlapping executions of scheduled task, so volatile will also work in this particular case even with increment.






share|improve this answer


























  • @BoristheSpider taking into account how scheduleWithFixedDelay work there will be no overlapped calls to counter++ in that particular scenario. So volatile should be OK.

    – Ivan
    yesterday













  • There is a happens-before relation introduced between subsequent execution of a task scheduled with scheduleWithFixedDelay.

    – Sotirios Delimanolis
    yesterday











  • How come volatile is not enough? The semantics of volatile should hold in every case and it would solve this problem. In the past, the writing of doubles and longs could have been more difficult to do atomically since those occupied more than one 32-bit register. However, as far as I can tell, that should not affect the semantics of volatile. Or am I missing something?

    – Edwin Dalorzo
    yesterday






  • 1





    @EdwinDalorzo in this particular case volatile is enough. But in general case not because increment is effectively translated into 3 operations: read, add, write and is prone to race condition in multithreaded environment

    – Ivan
    yesterday











  • To reiterate, the code they have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday
















6














No, this code is not thread-safe because there isn't any happens-before relation between increments made in different threads started with ScheduledExecutorService.



To fix it, you need to either mark the variable as volatile or switch to AtomicInteger or AtomicLong.



UPDATE:



As @BoristheSpider mentioned, in general in case of increment/decrement making a variable volatile is not enough since increment/decrement is not atomic itself and calling it from several threads concurrently will cause race conditions and missed updates. However, in this particular case scheduleWithFixedDelay() guarantees (according to Javadoc) that there will be overlapping executions of scheduled task, so volatile will also work in this particular case even with increment.






share|improve this answer


























  • @BoristheSpider taking into account how scheduleWithFixedDelay work there will be no overlapped calls to counter++ in that particular scenario. So volatile should be OK.

    – Ivan
    yesterday













  • There is a happens-before relation introduced between subsequent execution of a task scheduled with scheduleWithFixedDelay.

    – Sotirios Delimanolis
    yesterday











  • How come volatile is not enough? The semantics of volatile should hold in every case and it would solve this problem. In the past, the writing of doubles and longs could have been more difficult to do atomically since those occupied more than one 32-bit register. However, as far as I can tell, that should not affect the semantics of volatile. Or am I missing something?

    – Edwin Dalorzo
    yesterday






  • 1





    @EdwinDalorzo in this particular case volatile is enough. But in general case not because increment is effectively translated into 3 operations: read, add, write and is prone to race condition in multithreaded environment

    – Ivan
    yesterday











  • To reiterate, the code they have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday














6












6








6







No, this code is not thread-safe because there isn't any happens-before relation between increments made in different threads started with ScheduledExecutorService.



To fix it, you need to either mark the variable as volatile or switch to AtomicInteger or AtomicLong.



UPDATE:



As @BoristheSpider mentioned, in general in case of increment/decrement making a variable volatile is not enough since increment/decrement is not atomic itself and calling it from several threads concurrently will cause race conditions and missed updates. However, in this particular case scheduleWithFixedDelay() guarantees (according to Javadoc) that there will be overlapping executions of scheduled task, so volatile will also work in this particular case even with increment.






share|improve this answer















No, this code is not thread-safe because there isn't any happens-before relation between increments made in different threads started with ScheduledExecutorService.



To fix it, you need to either mark the variable as volatile or switch to AtomicInteger or AtomicLong.



UPDATE:



As @BoristheSpider mentioned, in general in case of increment/decrement making a variable volatile is not enough since increment/decrement is not atomic itself and calling it from several threads concurrently will cause race conditions and missed updates. However, in this particular case scheduleWithFixedDelay() guarantees (according to Javadoc) that there will be overlapping executions of scheduled task, so volatile will also work in this particular case even with increment.







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday









Peter Mortensen

13.9k1987113




13.9k1987113










answered yesterday









IvanIvan

5,71411022




5,71411022













  • @BoristheSpider taking into account how scheduleWithFixedDelay work there will be no overlapped calls to counter++ in that particular scenario. So volatile should be OK.

    – Ivan
    yesterday













  • There is a happens-before relation introduced between subsequent execution of a task scheduled with scheduleWithFixedDelay.

    – Sotirios Delimanolis
    yesterday











  • How come volatile is not enough? The semantics of volatile should hold in every case and it would solve this problem. In the past, the writing of doubles and longs could have been more difficult to do atomically since those occupied more than one 32-bit register. However, as far as I can tell, that should not affect the semantics of volatile. Or am I missing something?

    – Edwin Dalorzo
    yesterday






  • 1





    @EdwinDalorzo in this particular case volatile is enough. But in general case not because increment is effectively translated into 3 operations: read, add, write and is prone to race condition in multithreaded environment

    – Ivan
    yesterday











  • To reiterate, the code they have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday



















  • @BoristheSpider taking into account how scheduleWithFixedDelay work there will be no overlapped calls to counter++ in that particular scenario. So volatile should be OK.

    – Ivan
    yesterday













  • There is a happens-before relation introduced between subsequent execution of a task scheduled with scheduleWithFixedDelay.

    – Sotirios Delimanolis
    yesterday











  • How come volatile is not enough? The semantics of volatile should hold in every case and it would solve this problem. In the past, the writing of doubles and longs could have been more difficult to do atomically since those occupied more than one 32-bit register. However, as far as I can tell, that should not affect the semantics of volatile. Or am I missing something?

    – Edwin Dalorzo
    yesterday






  • 1





    @EdwinDalorzo in this particular case volatile is enough. But in general case not because increment is effectively translated into 3 operations: read, add, write and is prone to race condition in multithreaded environment

    – Ivan
    yesterday











  • To reiterate, the code they have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday

















@BoristheSpider taking into account how scheduleWithFixedDelay work there will be no overlapped calls to counter++ in that particular scenario. So volatile should be OK.

– Ivan
yesterday







@BoristheSpider taking into account how scheduleWithFixedDelay work there will be no overlapped calls to counter++ in that particular scenario. So volatile should be OK.

– Ivan
yesterday















There is a happens-before relation introduced between subsequent execution of a task scheduled with scheduleWithFixedDelay.

– Sotirios Delimanolis
yesterday





There is a happens-before relation introduced between subsequent execution of a task scheduled with scheduleWithFixedDelay.

– Sotirios Delimanolis
yesterday













How come volatile is not enough? The semantics of volatile should hold in every case and it would solve this problem. In the past, the writing of doubles and longs could have been more difficult to do atomically since those occupied more than one 32-bit register. However, as far as I can tell, that should not affect the semantics of volatile. Or am I missing something?

– Edwin Dalorzo
yesterday





How come volatile is not enough? The semantics of volatile should hold in every case and it would solve this problem. In the past, the writing of doubles and longs could have been more difficult to do atomically since those occupied more than one 32-bit register. However, as far as I can tell, that should not affect the semantics of volatile. Or am I missing something?

– Edwin Dalorzo
yesterday




1




1





@EdwinDalorzo in this particular case volatile is enough. But in general case not because increment is effectively translated into 3 operations: read, add, write and is prone to race condition in multithreaded environment

– Ivan
yesterday





@EdwinDalorzo in this particular case volatile is enough. But in general case not because increment is effectively translated into 3 operations: read, add, write and is prone to race condition in multithreaded environment

– Ivan
yesterday













To reiterate, the code they have posted is thread-safe. This answer is wrong.

– Sotirios Delimanolis
yesterday





To reiterate, the code they have posted is thread-safe. This answer is wrong.

– Sotirios Delimanolis
yesterday











2














No, this code is not thread-safe since there isn't any happens before relation between different threads accessing count.






share|improve this answer





















  • 1





    Even then it's not thread safe, because ++ isn't atomic.

    – Andy Turner
    yesterday






  • 1





    @michid please correct me if I am wrong but shouldn't counter++ be synchronized too as increment operation is not atomic.

    – Yug Singh
    yesterday






  • 1





    Does counter++ need to be synchronized even if it's guaranteed that only one thread will be running this runnable at a time (like in the code in the question)?

    – RandomQuestion
    yesterday













  • @RandomQuestion yes. Because visibility.

    – Boris the Spider
    yesterday








  • 1





    @RandomQuestion To reiterate, the code you have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday
















2














No, this code is not thread-safe since there isn't any happens before relation between different threads accessing count.






share|improve this answer





















  • 1





    Even then it's not thread safe, because ++ isn't atomic.

    – Andy Turner
    yesterday






  • 1





    @michid please correct me if I am wrong but shouldn't counter++ be synchronized too as increment operation is not atomic.

    – Yug Singh
    yesterday






  • 1





    Does counter++ need to be synchronized even if it's guaranteed that only one thread will be running this runnable at a time (like in the code in the question)?

    – RandomQuestion
    yesterday













  • @RandomQuestion yes. Because visibility.

    – Boris the Spider
    yesterday








  • 1





    @RandomQuestion To reiterate, the code you have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday














2












2








2







No, this code is not thread-safe since there isn't any happens before relation between different threads accessing count.






share|improve this answer















No, this code is not thread-safe since there isn't any happens before relation between different threads accessing count.







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday









Peter Mortensen

13.9k1987113




13.9k1987113










answered yesterday









michidmichid

5,54921938




5,54921938








  • 1





    Even then it's not thread safe, because ++ isn't atomic.

    – Andy Turner
    yesterday






  • 1





    @michid please correct me if I am wrong but shouldn't counter++ be synchronized too as increment operation is not atomic.

    – Yug Singh
    yesterday






  • 1





    Does counter++ need to be synchronized even if it's guaranteed that only one thread will be running this runnable at a time (like in the code in the question)?

    – RandomQuestion
    yesterday













  • @RandomQuestion yes. Because visibility.

    – Boris the Spider
    yesterday








  • 1





    @RandomQuestion To reiterate, the code you have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday














  • 1





    Even then it's not thread safe, because ++ isn't atomic.

    – Andy Turner
    yesterday






  • 1





    @michid please correct me if I am wrong but shouldn't counter++ be synchronized too as increment operation is not atomic.

    – Yug Singh
    yesterday






  • 1





    Does counter++ need to be synchronized even if it's guaranteed that only one thread will be running this runnable at a time (like in the code in the question)?

    – RandomQuestion
    yesterday













  • @RandomQuestion yes. Because visibility.

    – Boris the Spider
    yesterday








  • 1





    @RandomQuestion To reiterate, the code you have posted is thread-safe. This answer is wrong.

    – Sotirios Delimanolis
    yesterday








1




1





Even then it's not thread safe, because ++ isn't atomic.

– Andy Turner
yesterday





Even then it's not thread safe, because ++ isn't atomic.

– Andy Turner
yesterday




1




1





@michid please correct me if I am wrong but shouldn't counter++ be synchronized too as increment operation is not atomic.

– Yug Singh
yesterday





@michid please correct me if I am wrong but shouldn't counter++ be synchronized too as increment operation is not atomic.

– Yug Singh
yesterday




1




1





Does counter++ need to be synchronized even if it's guaranteed that only one thread will be running this runnable at a time (like in the code in the question)?

– RandomQuestion
yesterday







Does counter++ need to be synchronized even if it's guaranteed that only one thread will be running this runnable at a time (like in the code in the question)?

– RandomQuestion
yesterday















@RandomQuestion yes. Because visibility.

– Boris the Spider
yesterday







@RandomQuestion yes. Because visibility.

– Boris the Spider
yesterday






1




1





@RandomQuestion To reiterate, the code you have posted is thread-safe. This answer is wrong.

– Sotirios Delimanolis
yesterday





@RandomQuestion To reiterate, the code you have posted is thread-safe. This answer is wrong.

– Sotirios Delimanolis
yesterday


















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