What happens when you change a reference inside a method? (7)

Question

What happens when you change a reference inside a method?

public void reverseX(int[] nums) {
    int[] nums2 = new int[nums.length] ;
    for( int i=0 ; i < nums.length ; i++ )
        nums2[i] = nums[nums.length-(i+1)] ;
    nums = nums2 ;
};

This compiles fine.

Code is from here, and shown as an example of what not to do.

http://www.cs.nyu.edu/~cconway/teaching/cs1007/notes/arrays.pdf

It's not really hard to try and see right ? since the pointer to the array is passed by value - nothing will happen! — Nir Alfasi, Sep 15 '13 at 16:38
references are dynamic aliases. You make another array and make two pointers to it. — Val, Sep 15 '13 at 17:22
http://stackoverflow.com/questions/40480/is-java-pass-by-reference — nativist.bill.cutting, Sep 15 '13 at 17:33

score 4 · Accepted Answer · answered Sep 15 '13 at 16:37

Objects, reference types are suppose to be immutable.

Not all of them. Some, but not all. For example, String is immutable, but StringBuilder isn't. All arrays are mutable.

What happens here when nums is set to nums2.

The nums variable is given a new value, which is equal to the existing value of nums2.

That doesn't change the state of the objects that those variables refer to at all. It just changes the variable value itself.

Imagine the variables are pieces of paper. The integer array is like a house, and the values of the variables are like addresses written on the pieces of paper. This line:

nums = nums2 ;

... is just like copying what's written on one piece of paper onto another. That doesn't change the contents of the house that the address happens to refer to, right?

Now nums is just a local variable - because it's a parameter. That doesn't change anything in the calling code at all, because arguments are always passed by value in Java. So you're setting a local variable value just before the end of the method, and that's all. It will have no meaningful effect.

If none of this helps you, you'll need to ask a more specific question.

¹ Note that I'm not saying that a reference is necessarily a memory address - the word "address" is just convenient in the analogy.

the professor made a strange comment about arrays references being immutable when passed in as a parameter. This makes 0 sense, b.c. Java is always pass by Value and you are always working with a copy of what was passed in ... so I had to update my question. — nativist.bill.cutting, Sep 23 '13 at 14:24
@nativist.bill.cutting: Arrays themselves are always mutable. As for the references themselves - that doesn't make much sense as a concept, as you don't try to mutate a reference - you change the value of a variable, or data within an object. — Jon Skeet, Sep 23 '13 at 14:30

score 3 · Answer 2 · answered Sep 15 '13 at 16:37

3

Nothing happens. Reference types are passed by-value, so nums is a local variable. It doesn't affect anything outside the method.

answered Sep 15 '13 at 16:37

Oliver Charlesworth

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What happens when you change a reference inside a method? (7)

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