Auto optimisation for L cache for object's variables?
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
add a comment |
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
add a comment |
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
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
java multithreading volatile cpu-cache
edited Nov 12 '18 at 14:07
Andremoniy
asked Nov 12 '18 at 13:50
AndremoniyAndremoniy
21.8k674158
21.8k674158
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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.
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
add a comment |
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.
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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