Tuesday, 11 April 2017

What do Java objects look like in memory during run-time?

We know functions are implemented in memory as a stack of activation records. And we know Java methods are implemented as a stack of frames in JVM Stack and Java objects are allocated in Heap.

How do Java objects look like in heap? Once an object is laid out in memory, it's just a series of bytes.

Then how do we know where to look to find a particular field? Keep an internal table inside the compiler containing the offsets of each field.

Here is an example of an object layout for class "Base"(B). This class does not have any method, how methods are laid out in memory is in the next section.

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If we have another class "Derived"(D) extending this "Base" class. The memory layout would look like the following:

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Child object has the same memory layout as parent objects, except that it needs more space to place the newly added fields. The benefit of this layout is that A pointer of type B pointing at a D object still sees the B object at the beginning. Therefore, operations done on a D object through the B reference guaranteed to be safe, and there is no need to check what B points at dynamically.

Following the same logic, the method can be put at the beginning of the objects.

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However, this approach is not efficient. If a class has many methods(e.g. M), then each object must have O(M) pointers set. In addition, each object needs to have space for O(M) pointers. Those make creating objects slower and objects bigger.

The optimization approach is to create a virtual function table (or vtable) which is an array of pointers to the member function implementations for a particular class. Create a single instance of the vtable for each class and make each object store a pointer to the vtable.

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This is the optimized approach.