Calling a method and reference to an array

Question

public class midterm2 {
    static void methodOne(int[] a) {
        int[] b = new int[5];
        a=b;
        System.out.print(a.length);
        System.out.print(b.length);
    }
    public static void main(String[] args) {
        int[] a = new int[10];
        methodOne(a);
        System.out.print(a.length);

    }
}

The answer is 5510, and I don't understand why because I thought it would be 555. I thought the original array will be changed in this case.

Can anyone help me to understand this??

Thank you!

Milinda Arambawela · Accepted Answer · 2018-04-10 03:55:40Z

1

In main() method, a[] refers to int[10] array. When you passing int[10] array to methodOne(), new a[] variable creates for refer int[10] array and that variable is a[]. Now you have 2 a[] variables which refers int[10] array.

Now you create b[] and it refers int[5] array. You assign b[]'s array to a[] (which belongs to methodOne()). but a[] in main method still refers int[10] array.

answered Apr 10, 2018 at 3:55

Milinda Arambawela

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Comments

Mohit Mutha · Accepted Answer · 2018-04-10 03:49:59Z

1

Since a is a array of primitive type it is a pass by value to the methodOne() and hence the variable a in the scope of the main method remains unaltered.

answered Apr 10, 2018 at 3:49

Mohit Mutha

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Comments

hj.sun · Accepted Answer · 2018-04-10 03:50:23Z

1

change methodOne parameter name to "int[] c" will make you understand easily. Parameter "a" in methodOne is totally different with variable "a" in main method.

answered Apr 10, 2018 at 3:50

hj.sun

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Comments

nj2237 · Accepted Answer · 2018-04-10 03:53:05Z

1

a=b; statement assigns reference of "b" to "a".

If you code like a = new int[10]; it would print 5510 as output. So unless you use "new" it will act as call by value and changes made in the called method will not reflect in calling method.

edited Apr 10, 2018 at 3:53

nj2237

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answered Apr 10, 2018 at 3:49

Jayanand Raghuwanshi

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Comments

Sunil · Accepted Answer · 2018-04-10 03:53:52Z

1

The short answer is: The reference variable 'a' is local to functions 'main' and 'methodOne'.

answered Apr 10, 2018 at 3:53

Sunil

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Comments

MinA · Accepted Answer · 2018-04-10 04:37:34Z

1

In your: methodOne() 'a' is local to that method. so when you do a = b,

it doesnt do any change to varible 'a' in your main().

Then again you are going to print variable 'a' in main(): So it refers to local variable a an print it.

answered Apr 10, 2018 at 4:37

MinA

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