Question

I have here a simple question related to Java.
Let’s say you have an int array as instance variable:

int[] in = new int[5];

So, now by default it contains 5 zeros.
But what if you have the same array as local variable. Does it get initialized to zeros? That is not a homework, I am learning Java language.
Best regards

Answer 1

First thing to understand is that, local varibles are stored on stack which are not initialized explicitly with their default values. While instance variables are stored on Heap, and they are by default initialized with their default value.

Also, objects are also created on Heap, regardless of whether an instance reference variable is holding its reference, or a local reference variable.

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Now, what happens is, when you declare your array reference like this as local variable, and initialize it with an array: –

int[] in = new int[5];

The array reference (in) is stored on stack, and a memory is allocated for array capable of holding 5 integer elements on heap (Remember, objects are created on Heap). Then, 5 contiguous memory location (size = 5), for storing integer value are allocated on Heap. And each index on array object holds a reference to those memory location in sequence. Then the array reference points to that array. So, since memory for 5 integer values are allocated on Heap, they are initialized to their default value.

And also, when you declare your array reference, and don’t initialize it with any array object: –

int[] in;

The array reference is created on Stack (as it is a local variable), but it does not gets initialized to an array by default, and neither to null, as is the case with instance variables.

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So, this is how allocation looks like when you use the first way of array declaration and initialization: –

"Your array reference"
     "on stack"    

       |    |          "Array object on Heap"
       +----+                  
       | in |---------->  ([0, 0, 0, 0, 0])
       +----+
       "Stack"                  "Heap"