Java中的方法的可变范围

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...

Answer 1

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.

Answer 2

您可能需要阅读有关Java中的值传递的更多信息。

Answer 3

Just may be to make it clear, All variables passed by value means that the called method just gets the values of parameters and not a reference (pointer) to these objects, so any modification on these objects in the method body doesn't affect the outer object. Unlike c++ which has the option to pass the object by reference where the method gets the actual object not it's value, so any modification in the method body affects the outer object.Java doesn't have pass by reference.

Answer 4

在Java中，变量按值传递。