Question

Consider:

public SomeClass implements SomeInterface{...}

SomeClass obj = new SomeClass();

SomeInterface x = obj;

I am trying to relate line 3 to my very basic understanding of memory management.
I know the memory location represented by “obj” just contains a pointer to the memory location of SomeClass. Assuming I am using a 64bit JVM, then up to 64 bits are allocated for the “obj” pointer. What is created in memory when the JRE implements x? Is it just a 64bit pointer to SomeClass?

Answer 1

Every object reference takes up the same amount of memory, no matter how you declare it.

So x and obj are two distict references, which just happen to point at the same thing.