Java浅拷贝数组

Question

if this is a shallow copy

double[] a = new double[100];
a = b; // let b be some other double array[100]

I understand that the better way to do this is using a for loop or using

System.arrayCopy(b,0,a,0,100);

However what happens to this?

public double[] function1(){
    returns somedouble[100];
}

double[] a = new double[100];
a = function1(); // i believe this will also be a shallow copy

System.arrayCopy(function1(),0,a,0,100); // will this call function1 100 times?

Answer 1

double[] a = new double[100];
a = b; // let b be some other double array[100]

Create an array named a of double with size of 100. Now when you a = b will copy the reference of b array to variable a.

     +--------------------------------------------+  <- Suppose whole
a->  | 2.5 |  |  |  |  |  |  |  |  |  |  |  |  |  |  array is filled 
     +--------------------------------------------+  with value 2.5

     +--------------------------------------------+  <- Suppose whole
b->  | 7.9 |  |  |  |  |  |  |  |  |  |  |  |  |  |  array is filled 
     +--------------------------------------------+  with value 7.9

after a = b

     +--------------------------------------------+  <- Suppose whole
     | 2.5 |  |  |  |  |  |  |  |  |  |  |  |  |  |  array is filled 
     +--------------------------------------------+  with value 2.5

a->  +--------------------------------------------+  <- Suppose whole
b->  | 7.9 |  |  |  |  |  |  |  |  |  |  |  |  |  |  array is filled 
     +--------------------------------------------+  with value 7.9

So now a and b pointing to same array.

public double[] function1(){
    return somedouble[100];
}

double[] a = new double[100];
a = function1();

Now same thing is happening here. You create an array named a and then call function1() and again assigned the returning array reference to a.

System.arraycopy(function1(), 0, a, 0, 100);

Here the calling order will be

1 -> function1() will be called and the returning array reference will be saved in a temporary variable.

2 -> call to System.arraycopy(temporary variable, 0, a, 0, 100)

So function1() will be called only once.

As a side note make sure you use System.arraycopy(args) instead of System.arrayCopy(args)

Answer 2

    double[] a = new double[100];
    a = b; // let b be some other double array[100]

First, It's assigning, not copying.

---

    double[] a = new double[100];
    a = function1(); // i believe this will also be a shallow copy

It's assigning. you assign the returned value somedouble[100] to a

---

System.arrayCopy(function1(),0,a,0,100); // will this call function1 100 times?

No, it won't call function1 100 times. The above code mostly equals to

    double[] tmpref = function1();
    System.arrayCopy(tmparr,0,a,0,100);

Because it first evaluates the arguments, then it calls arrayCopy