Auto optimisation for L cache for object's variables?










2















Frankly, this is a continue of this my question, inspired by this answer: https://stackoverflow.com/a/53262717/1479414



Let's suppose we have a class:



public class Foo 
private Integer x;

public void setX(Integer x)
this.x = x;


public Integer getX()
return this.x;




And let us consider a very specific scenario, when we have just two threads which interact with the x variable:



At time 1, a thread T1 is created



At time 2, T1 sets the value: foo.setX(123);



At time 3, a thread T2 is created



At time 4, T2 reads the value: foo.getX();



No other threads interact with this value. Both operations happen only once.



So there is no apparent x value read operations before the thread T2 does his job.



The question is: does any auto optimisation for L cache exist which can cache null value of the x variable, so the thread T2 will read its cached value?
In other words, do we need the volatile modifier in this particular scenario?










share|improve this question




























    2















    Frankly, this is a continue of this my question, inspired by this answer: https://stackoverflow.com/a/53262717/1479414



    Let's suppose we have a class:



    public class Foo 
    private Integer x;

    public void setX(Integer x)
    this.x = x;


    public Integer getX()
    return this.x;




    And let us consider a very specific scenario, when we have just two threads which interact with the x variable:



    At time 1, a thread T1 is created



    At time 2, T1 sets the value: foo.setX(123);



    At time 3, a thread T2 is created



    At time 4, T2 reads the value: foo.getX();



    No other threads interact with this value. Both operations happen only once.



    So there is no apparent x value read operations before the thread T2 does his job.



    The question is: does any auto optimisation for L cache exist which can cache null value of the x variable, so the thread T2 will read its cached value?
    In other words, do we need the volatile modifier in this particular scenario?










    share|improve this question


























      2












      2








      2








      Frankly, this is a continue of this my question, inspired by this answer: https://stackoverflow.com/a/53262717/1479414



      Let's suppose we have a class:



      public class Foo 
      private Integer x;

      public void setX(Integer x)
      this.x = x;


      public Integer getX()
      return this.x;




      And let us consider a very specific scenario, when we have just two threads which interact with the x variable:



      At time 1, a thread T1 is created



      At time 2, T1 sets the value: foo.setX(123);



      At time 3, a thread T2 is created



      At time 4, T2 reads the value: foo.getX();



      No other threads interact with this value. Both operations happen only once.



      So there is no apparent x value read operations before the thread T2 does his job.



      The question is: does any auto optimisation for L cache exist which can cache null value of the x variable, so the thread T2 will read its cached value?
      In other words, do we need the volatile modifier in this particular scenario?










      share|improve this question
















      Frankly, this is a continue of this my question, inspired by this answer: https://stackoverflow.com/a/53262717/1479414



      Let's suppose we have a class:



      public class Foo 
      private Integer x;

      public void setX(Integer x)
      this.x = x;


      public Integer getX()
      return this.x;




      And let us consider a very specific scenario, when we have just two threads which interact with the x variable:



      At time 1, a thread T1 is created



      At time 2, T1 sets the value: foo.setX(123);



      At time 3, a thread T2 is created



      At time 4, T2 reads the value: foo.getX();



      No other threads interact with this value. Both operations happen only once.



      So there is no apparent x value read operations before the thread T2 does his job.



      The question is: does any auto optimisation for L cache exist which can cache null value of the x variable, so the thread T2 will read its cached value?
      In other words, do we need the volatile modifier in this particular scenario?







      java multithreading volatile cpu-cache






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 12 '18 at 14:07







      Andremoniy

















      asked Nov 12 '18 at 13:50









      AndremoniyAndremoniy

      21.8k674158




      21.8k674158






















          2 Answers
          2






          active

          oldest

          votes


















          2














          When you create a thread, it will see any value set before it is created.



          In the Javadoc for java.util.concurrency for Java 11 under Memory Visibility Properties and JLS-17.4.5 for Java 8 states




          A call to start on a thread happens-before any action in the started thread.




          NOTE: A thread only reads a cached value, when it re-reads a value in a cache line it has already in its cache. If it reads a cache line it has never read before or is no longer in a cache, it won't read a stale value.






          share|improve this answer

























          • Thank you a lot for your answer! It sounds logical that when a new thread is created it should see the actualised value of any variable. However, could this statement be strengthened with some links to the specification?

            – Andremoniy
            Nov 12 '18 at 19:03






          • 1





            @Andremoniy I have added a quote.

            – Peter Lawrey
            Nov 12 '18 at 19:19


















          1














          T1 and T2 are executed sequentially and cache is coherent, especially in this sequential use case.



          So there is no way that T2 at time 4 will get a null value.






          share|improve this answer























          • So you assume that the CPU cache has nothing to do with volatile-problem?

            – Andremoniy
            Nov 12 '18 at 15:12












          • I do not assume that. I assume that because the execution of T1 and T2 is sequential, the cache will be coherent (no concurrency issue)

            – Nicolas Labrot
            Nov 12 '18 at 15:31











          • That was my question: some people claim that there exist an auto optimisaiton which acts independently of existent threads

            – Andremoniy
            Nov 12 '18 at 15:37






          • 1





            Could you please elaborate?

            – Nicolas Labrot
            Nov 12 '18 at 16:31











          • I don't know how to elaborate. It was said that there is a "auto optimisation for L* cache" which can cause the null value be cached and passed therefore to the thread T2. That was my question: does something similar really exist?

            – Andremoniy
            Nov 12 '18 at 19:04










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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          When you create a thread, it will see any value set before it is created.



          In the Javadoc for java.util.concurrency for Java 11 under Memory Visibility Properties and JLS-17.4.5 for Java 8 states




          A call to start on a thread happens-before any action in the started thread.




          NOTE: A thread only reads a cached value, when it re-reads a value in a cache line it has already in its cache. If it reads a cache line it has never read before or is no longer in a cache, it won't read a stale value.






          share|improve this answer

























          • Thank you a lot for your answer! It sounds logical that when a new thread is created it should see the actualised value of any variable. However, could this statement be strengthened with some links to the specification?

            – Andremoniy
            Nov 12 '18 at 19:03






          • 1





            @Andremoniy I have added a quote.

            – Peter Lawrey
            Nov 12 '18 at 19:19















          2














          When you create a thread, it will see any value set before it is created.



          In the Javadoc for java.util.concurrency for Java 11 under Memory Visibility Properties and JLS-17.4.5 for Java 8 states




          A call to start on a thread happens-before any action in the started thread.




          NOTE: A thread only reads a cached value, when it re-reads a value in a cache line it has already in its cache. If it reads a cache line it has never read before or is no longer in a cache, it won't read a stale value.






          share|improve this answer

























          • Thank you a lot for your answer! It sounds logical that when a new thread is created it should see the actualised value of any variable. However, could this statement be strengthened with some links to the specification?

            – Andremoniy
            Nov 12 '18 at 19:03






          • 1





            @Andremoniy I have added a quote.

            – Peter Lawrey
            Nov 12 '18 at 19:19













          2












          2








          2







          When you create a thread, it will see any value set before it is created.



          In the Javadoc for java.util.concurrency for Java 11 under Memory Visibility Properties and JLS-17.4.5 for Java 8 states




          A call to start on a thread happens-before any action in the started thread.




          NOTE: A thread only reads a cached value, when it re-reads a value in a cache line it has already in its cache. If it reads a cache line it has never read before or is no longer in a cache, it won't read a stale value.






          share|improve this answer















          When you create a thread, it will see any value set before it is created.



          In the Javadoc for java.util.concurrency for Java 11 under Memory Visibility Properties and JLS-17.4.5 for Java 8 states




          A call to start on a thread happens-before any action in the started thread.




          NOTE: A thread only reads a cached value, when it re-reads a value in a cache line it has already in its cache. If it reads a cache line it has never read before or is no longer in a cache, it won't read a stale value.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 12 '18 at 19:19

























          answered Nov 12 '18 at 18:39









          Peter LawreyPeter Lawrey

          442k56559965




          442k56559965












          • Thank you a lot for your answer! It sounds logical that when a new thread is created it should see the actualised value of any variable. However, could this statement be strengthened with some links to the specification?

            – Andremoniy
            Nov 12 '18 at 19:03






          • 1





            @Andremoniy I have added a quote.

            – Peter Lawrey
            Nov 12 '18 at 19:19

















          • Thank you a lot for your answer! It sounds logical that when a new thread is created it should see the actualised value of any variable. However, could this statement be strengthened with some links to the specification?

            – Andremoniy
            Nov 12 '18 at 19:03






          • 1





            @Andremoniy I have added a quote.

            – Peter Lawrey
            Nov 12 '18 at 19:19
















          Thank you a lot for your answer! It sounds logical that when a new thread is created it should see the actualised value of any variable. However, could this statement be strengthened with some links to the specification?

          – Andremoniy
          Nov 12 '18 at 19:03





          Thank you a lot for your answer! It sounds logical that when a new thread is created it should see the actualised value of any variable. However, could this statement be strengthened with some links to the specification?

          – Andremoniy
          Nov 12 '18 at 19:03




          1




          1





          @Andremoniy I have added a quote.

          – Peter Lawrey
          Nov 12 '18 at 19:19





          @Andremoniy I have added a quote.

          – Peter Lawrey
          Nov 12 '18 at 19:19













          1














          T1 and T2 are executed sequentially and cache is coherent, especially in this sequential use case.



          So there is no way that T2 at time 4 will get a null value.






          share|improve this answer























          • So you assume that the CPU cache has nothing to do with volatile-problem?

            – Andremoniy
            Nov 12 '18 at 15:12












          • I do not assume that. I assume that because the execution of T1 and T2 is sequential, the cache will be coherent (no concurrency issue)

            – Nicolas Labrot
            Nov 12 '18 at 15:31











          • That was my question: some people claim that there exist an auto optimisaiton which acts independently of existent threads

            – Andremoniy
            Nov 12 '18 at 15:37






          • 1





            Could you please elaborate?

            – Nicolas Labrot
            Nov 12 '18 at 16:31











          • I don't know how to elaborate. It was said that there is a "auto optimisation for L* cache" which can cause the null value be cached and passed therefore to the thread T2. That was my question: does something similar really exist?

            – Andremoniy
            Nov 12 '18 at 19:04















          1














          T1 and T2 are executed sequentially and cache is coherent, especially in this sequential use case.



          So there is no way that T2 at time 4 will get a null value.






          share|improve this answer























          • So you assume that the CPU cache has nothing to do with volatile-problem?

            – Andremoniy
            Nov 12 '18 at 15:12












          • I do not assume that. I assume that because the execution of T1 and T2 is sequential, the cache will be coherent (no concurrency issue)

            – Nicolas Labrot
            Nov 12 '18 at 15:31











          • That was my question: some people claim that there exist an auto optimisaiton which acts independently of existent threads

            – Andremoniy
            Nov 12 '18 at 15:37






          • 1





            Could you please elaborate?

            – Nicolas Labrot
            Nov 12 '18 at 16:31











          • I don't know how to elaborate. It was said that there is a "auto optimisation for L* cache" which can cause the null value be cached and passed therefore to the thread T2. That was my question: does something similar really exist?

            – Andremoniy
            Nov 12 '18 at 19:04













          1












          1








          1







          T1 and T2 are executed sequentially and cache is coherent, especially in this sequential use case.



          So there is no way that T2 at time 4 will get a null value.






          share|improve this answer













          T1 and T2 are executed sequentially and cache is coherent, especially in this sequential use case.



          So there is no way that T2 at time 4 will get a null value.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 12 '18 at 14:38









          Nicolas LabrotNicolas Labrot

          2,8621828




          2,8621828












          • So you assume that the CPU cache has nothing to do with volatile-problem?

            – Andremoniy
            Nov 12 '18 at 15:12












          • I do not assume that. I assume that because the execution of T1 and T2 is sequential, the cache will be coherent (no concurrency issue)

            – Nicolas Labrot
            Nov 12 '18 at 15:31











          • That was my question: some people claim that there exist an auto optimisaiton which acts independently of existent threads

            – Andremoniy
            Nov 12 '18 at 15:37






          • 1





            Could you please elaborate?

            – Nicolas Labrot
            Nov 12 '18 at 16:31











          • I don't know how to elaborate. It was said that there is a "auto optimisation for L* cache" which can cause the null value be cached and passed therefore to the thread T2. That was my question: does something similar really exist?

            – Andremoniy
            Nov 12 '18 at 19:04

















          • So you assume that the CPU cache has nothing to do with volatile-problem?

            – Andremoniy
            Nov 12 '18 at 15:12












          • I do not assume that. I assume that because the execution of T1 and T2 is sequential, the cache will be coherent (no concurrency issue)

            – Nicolas Labrot
            Nov 12 '18 at 15:31











          • That was my question: some people claim that there exist an auto optimisaiton which acts independently of existent threads

            – Andremoniy
            Nov 12 '18 at 15:37






          • 1





            Could you please elaborate?

            – Nicolas Labrot
            Nov 12 '18 at 16:31











          • I don't know how to elaborate. It was said that there is a "auto optimisation for L* cache" which can cause the null value be cached and passed therefore to the thread T2. That was my question: does something similar really exist?

            – Andremoniy
            Nov 12 '18 at 19:04
















          So you assume that the CPU cache has nothing to do with volatile-problem?

          – Andremoniy
          Nov 12 '18 at 15:12






          So you assume that the CPU cache has nothing to do with volatile-problem?

          – Andremoniy
          Nov 12 '18 at 15:12














          I do not assume that. I assume that because the execution of T1 and T2 is sequential, the cache will be coherent (no concurrency issue)

          – Nicolas Labrot
          Nov 12 '18 at 15:31





          I do not assume that. I assume that because the execution of T1 and T2 is sequential, the cache will be coherent (no concurrency issue)

          – Nicolas Labrot
          Nov 12 '18 at 15:31













          That was my question: some people claim that there exist an auto optimisaiton which acts independently of existent threads

          – Andremoniy
          Nov 12 '18 at 15:37





          That was my question: some people claim that there exist an auto optimisaiton which acts independently of existent threads

          – Andremoniy
          Nov 12 '18 at 15:37




          1




          1





          Could you please elaborate?

          – Nicolas Labrot
          Nov 12 '18 at 16:31





          Could you please elaborate?

          – Nicolas Labrot
          Nov 12 '18 at 16:31













          I don't know how to elaborate. It was said that there is a "auto optimisation for L* cache" which can cause the null value be cached and passed therefore to the thread T2. That was my question: does something similar really exist?

          – Andremoniy
          Nov 12 '18 at 19:04





          I don't know how to elaborate. It was said that there is a "auto optimisation for L* cache" which can cause the null value be cached and passed therefore to the thread T2. That was my question: does something similar really exist?

          – Andremoniy
          Nov 12 '18 at 19:04

















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