Java-多线程CPU缓存

Question

I know the threads save the values of the variables in the cpu cache where it is running because in this way the cpu doesnt have wait so much time when it's necessary to get the values inside in the variables.

But for example if i have this object

public class MyObject {
    int a = 2;
}

and now the thread do something like this:

MyObject obj = new MyObject();
obj.a=3;

My question is:
what will be saved in the cpu cache ? all the MyObject structure or just the reference? I think all the structure (have more sense) but i prefer to ask because i would like to be sure about that.

I'm a noob about multithread and i'm sure is more complex how a cpu cache works, but at the moment i need just basic information.

Answer 1

In your example, only one thread is acting. For this thread, cache is transparent - there is no way to determine if a value is in cache, in main memory, or both. First all values are put in the cache but then very soon, in an unknown moment of time they are pushed out.

"i would like to be sure about that" - why? Your program behaviour does not depend on this.

Answer 2

These question has two sides:

What the CPU is doing: The CPU is designed to keep everything in the cache that is needed very often. If you change a value it will keep changes in the cache until it is needs to write it to the main memory (actually it depends on the CPUs strategy write-back vs write-through). The "need" to write it to main memory is programatically controlled or the CPU descides its needing space for other stuff. To answer one part of your question: For the CPU everything is data, the value you set in Java, and the internal object data structures. To access your value, you need the object address first, so that is very probably in the cache, too :)

The second point, is what Java programmer should expect and not expect: This is very exactly defined in the Java Memory Model. Just start here: http://en.wikipedia.org/wiki/Java_Memory_Model

So for your lines:

MyObject obj = new MyObject();
obj.a=3;

There is no guarantee that another thread running after this code, sees the new value. And it also may not see your new object reference but null instead. You need a synchronized block or a volatile variable.