Variable scope with a method in Java

Question

I thought I understood variable scope until I came across this bit of code:

private static void someMethod(int i, Account a) {
  i++;
  a.deposit(5);
  a = new Account(80);
}

int score = 10;
Account account = new Account(100);
someMethod(score, account);
System.out.println(score); // prints 10
System.out.println(account.balance); // prints 105!!!

EDIT: I understand why a=new Account(80) would not do anything but I'm confused about a.deposit(5) actually working since a is just a copy of the original Account being passed in...

http://javadude.com/articles/passbyvalue.htm – helpermethod Oct 03 '10 at 08:36 — helpermethod, Oct 03 '10 at 08:36

score 6 · Accepted Answer · answered Oct 03 '10 at 02:20

The variable a is a copy of the reference being passed in, so it still has the same value and refers to the same Account object as the account variable (that is, until you reassign a). When you make the deposit, you're still working with a reference to the original object that is still referred to in the outer scope.

score 3 · Answer 2 · edited May 23 '17 at 12:15

3

It might be time for you to read more about pass-by-value in Java.

edited May 23 '17 at 12:15

Community

1
1

answered Oct 03 '10 at 02:10

matt b

138,234
66
282
345

score -2 · Answer 3 · answered Oct 03 '10 at 02:10

-2

In java, variables are passed by value.

answered Oct 03 '10 at 02:10

fastcodejava

39,895
28
133
186

I think you should remove this, since objects are passed by reference. – stacker Oct 08 '10 at 20:06
@stacker - how can that be? Just the answer before says it is pass by value. – fastcodejava Oct 08 '10 at 20:44
This post should be removed as it is completly wrong. – Kir Kanos Dec 19 '22 at 12:27

Variable scope with a method in Java

3 Answers3