Question

Based on the understanding from the following:

Where is allocated variable reference, in stack or in the heap?

I was wondering since all the objects are created on the common heap. If multiple threads create objects then to prevent data corruption there has to be some serialization that must be happening to prevent the multiple threads from creating objects at same locations. Now, with a large number of threads this serialization would cause a big bottleneck. How does Java avoid this bottleneck? Or am I missing something?

Any help appreciated.

Answer 1

Modern VM implementations reserve for each thread an own area on the heap to create objects in. So, no problem as long as this area does not get full (then the garbage collector moves the surviving objects).

Further read: how TLAB works in Sun’s JVM. Azul’s VM uses slightly different approach (look at “A new thread & stack layout”), the article shows quite a few tricks JVMs may perform behind the scenes to ensure nowadays Java speed.

The main idea is keeping per thread (non-shared) area to allocate new objects, much like allocating on the stack with C/C++. The copy garbage collection is very quick to deallocate the short-lived objects, the few survivors, if any, are moved into different area. Thus, creating relatively small objects is very fast and lock free.

The lock free allocation is very important, especially since the question regards multithreaded environment. It also allows true lock-free algorithms to exist. Even if an algorithm, itself, is a lock-free but allocation of new objects is synchronized, the entire algorithm is effectively synchronized and ultimately less scalable.
java.util.concurrent.ConcurrentLinkedQueue which is based on the work of Maged M. Michael Michael L. Scott is a classic example.

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What happens if an object is referenced by another thread? (due to discussion request)

That object (call it A) will be moved to some “survivor” area. The survivor area is checked less often than the ThreadLocal areas. It contains, like the name suggests, objects whose references managed to escape, or in particular A managed to stay alive. The copy (move) part occurs during some “safe point” (safe point excludes properly JIT’d code), so the garbage collector is sure the object is not being referenced. The references to the object are updated, the necessary memories fences issued and the application (java code) is free to continue. Further read to this simplistic scenario.

To the very interested reader and if possible to chew it: the highly advanced Pauseless GC Algorithm