Are arrays passed by value or passed by reference in Java? [duplicate]

Question

Everything in Java is passed by value. In case of an array (which is nothing but an Object), the array reference is passed by value (just like an object reference is passed by value).

When you pass an array to other method, actually the reference to that array is copied.

• Any changes in the content of array through that reference will affect the original array.
• But changing the reference to point to a new array will not change the existing reference in original method.

See this post: Is Java "pass-by-reference" or "pass-by-value"?

See this working example:

public static void changeContent(int[] arr) {

   // If we change the content of arr.
   arr[0] = 10;  // Will change the content of array in main()
}

public static void changeRef(int[] arr) {
   // If we change the reference
   arr = new int[2];  // Will not change the array in main()
   arr[0] = 15;
}

public static void main(String[] args) {
    int [] arr = new int[2];
    arr[0] = 4;
    arr[1] = 5;

    changeContent(arr);

    System.out.println(arr[0]);  // Will print 10.. 
  
    changeRef(arr);

    System.out.println(arr[0]);  // Will still print 10.. 
                                 // Change the reference doesn't reflect change here..
}

Your question is based on a false premise.

Arrays are not a primitive type in Java, but they are not objects either ... "

In fact, all arrays in Java are objects¹. Every Java array type has java.lang.Object as its supertype, and inherits the implementation of all methods in the Object API.

... so are they passed by value or by reference? Does it depend on what the array contains, for example references or a primitive type?

Short answers: 1) pass by value, and 2) it makes no difference.

Longer answer:

Like all Java objects, arrays are passed by value ... but the value is the reference to the array. So, when you assign something to a cell of the array in the called method, you will be assigning to the same array object that the caller sees.

This is NOT pass-by-reference. Real pass-by-reference involves passing the address of a variable. With real pass-by-reference, the called method can assign to its local variable, and this causes the variable in the caller to be updated.

But not in Java. In Java, the called method can update the contents of the array, and it can update its copy of the array reference, but it can't update the variable in the caller that holds the caller's array reference. Hence ... what Java is providing is NOT pass-by-reference.

Here are some links that explain the difference between pass-by-reference and pass-by-value. If you don't understand my explanations above, or if you feel inclined to disagree with the terminology, you should read them.

• http://publib.boulder.ibm.com/infocenter/comphelp/v8v101/topic/com.ibm.xlcpp8a.doc/language/ref/cplr233.htm
• http://www.cs.fsu.edu/~myers/c++/notes/references.html

Related SO question:

• Is Java "pass-by-reference" or "pass-by-value"?

Historical background:

The phrase "pass-by-reference" was originally "call-by-reference", and it was used to distinguish the argument passing semantics of FORTRAN (call-by-reference) from those of ALGOL-60 (call-by-value and call-by-name).

• In call-by-value, the argument expression is evaluated to a value, and that value is copied to the called method.

• In call-by-reference, the argument expression is partially evaluated to an "lvalue" (i.e. the address of a variable or array element) that is passed to the calling method. The calling method can then directly read and update the variable / element.

• In call-by-name, the actual argument expression is passed to the calling method (!!) which can evaluate it multiple times (!!!). This was complicated to implement, and could be used (abused) to write code that was very difficult to understand. Call-by-name was only ever used in Algol-60 (thankfully!).

UPDATE

Actually, Algol-60's call-by-name is similar to passing lambda expressions as parameters. The wrinkle is that these not-exactly-lambda-expressions (they were referred to as "thunks" at the implementation level) can indirectly modify the state of variables that are in scope in the calling procedure / function. That is part of what made them so hard to understand. (See the Wikipedia page on Jensen's Device for example.)

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^{1. Nothing in the linked Q&A (Arrays in Java and how they are stored in memory) either states or implies that arrays are not objects.}

Arrays are in fact objects, so a reference is passed (the reference itself is passed by value, confused yet?). Quick example:

// assuming you allocated the list
public void addItem(Integer[] list, int item) {
    list[1] = item;
}

You will see the changes to the list from the calling code. However you can't change the reference itself, since it's passed by value:

// assuming you allocated the list
public void changeArray(Integer[] list) {
    list = null;
}

If you pass a non-null list, it won't be null by the time the method returns.

No that is wrong. Arrays are special objects in Java. So it is like passing other objects where you pass the value of the reference, but not the reference itself. Meaning, changing the reference of an array in the called routine will not be reflected in the calling routine.