int数组初始化

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"

Answer 2

It is the same thing if you do :

int[] in = new int[5] as instance variable or local variable. The array object in will contain zeros in both cases.

Difference would be if you would do something like :

1. Instance variable : int[] in ; (it is initialized with null ), and the in object will live in heap space .

2. Local variable : int[] in ; (it has to be initialized by the user) will live in stack

Answer 3

For primitive type arrays it is initialized to their default values. In the documentation it says :

a single-dimensional array is created of the specified length, and each component of the array is initialized to its default value

For the integer type default value is 0.

Answer 4

Yes, when you initialise an array the contents will be set to the default value for that type, for int it would be 0 and for a reference type it would be null .

If you initialise an array and inspect the contents you can see this for yourself:

...
final int[] in = new int[5];

for (int i = 0; i < in.length; i++) {
    System.out.println(in[i]);
}
...

This will print:

0
0
0
0
0

Answer 5

yes

public void method() {
    int[] in = new int[5];
    System.out.pritnln(in[0]); //output in 0
}

In this case, your Array is a local variable , all you need is to initialize your array . once you initialize you array, voila your array element**s get their **default values .

Answer 6

It does not really matter whether the declared array in an instance variable or a local variable it will get initialized to the default value.

Each class variable, instance variable, or array component is initialized with a default value when it is created.

As per JLS

An array initializer creates an array and provides initial values for all its components.

Answer 7

将数组实例化为局部变量时，数组不包含5个零。