How do I tell the difference between a variable and a reference in Java?

Question

I'm coming from a background in C and C++, where I grew pretty comfortable with pointers and references. Now I'm learning Data Structures and Algorithms in Java, and I saw a slide talking about nodes in Linked Lists being structured as so

class Node<E>{
    E element;
    Node<E> next;
    public Node(E o){
        element = o;
    }
}

Given my background, to me this looks like the next node is contained directly within this node, and there's nothing to clarify that it's a reference (pointer?) to a separate address in memory. I also know at this point that Java doesn't have pass-by-reference at all, which makes this even weirder (from my perspective). So I want to know, how do I tell the difference between variables that hold actual values or objects, and variables that hold references to other values/objects

I think this https://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value explains well the characteristic of Java in this aspect. — Hung Thai, May 25 '22 at 04:14
In Java, it's always a reference (a "pointer"). There *IS* no way to declare a variable as the value / object itself. — markspace, May 25 '22 at 04:14
Side notes: primitives are obviously always themselves, never a reference. And while Java is pass by value, since objects are always represented by a reference you always get reference semantics when passing them to a method / subroutine. — markspace, May 25 '22 at 04:25

Michael M. · Answer 1 · 2022-05-25T04:22:02.950

In Java, unlike C or C++, all variables with types that are not primitives are references. In this case, next is a reference because Node is an object, not a primitive. element would also be a reference to another object, unless E is a primitive. Java's garbage collector handles all the memory management for these references, so you don't need to worry about allocation and freeing the memory for the underlying objects. For example:

Foo bar = new Foo();    // bar is an object of the class Foo
Foo foobar = bar;       // all objects are references, so bar and foobar are
                        // referencing the same underlying memory

int a = 1;              // a is an int
int b = a;              // the value of a is copied into b, because a is a 
                        // primitive

score 0 · Answer 2 · answered May 25 '22 at 05:07

In Java, variables fall into two categories.

primitives (int, short, boolean, char, long, byte, double, float),
references (everything else).

The type parameter E can't be a primitive type in this context, so in your code, element and next both store references to something. That's why your class succeeds in implementing a linked list.

As for the question of how this works with "pass by value" - argument values passed to a method are either primitives or references, just like the values of any other Java variable. So if your parameter is a reference type, not a primitive; then although you're doing "pass by value", it behaves a little like "pass by reference", insofar as the method being called now has access to the original object, not a copy. Of course, it's not really "pass by reference" because the method that was called can't change what the original variable refers to.

How do I tell the difference between a variable and a reference in Java?

2 Answers2