var中的第二个将如何为2，因为我们正在对其范围之外进行访问

Question

Please explain how the second in var will be 2 since, we are accessing outside it scope. o/p - in=1 out=2 in=2

class Test
{
    public static void main(String args[])
    {
        int var = 1;
        System.out.println("in="+var);
        {
            var = 2;
            System.out.println("out="+var);
        }
        System.out.println("in="+var);
    }
}

Answer 1

The scope of var is controlled by the outer declaration. You only have a single var . You can't shadow var as posted because it is a local variable. However , if we tweak it a little bit for the example.

static int var = 1;
public static void main(String args[])
{
    System.out.println("in="+var);
    {
        int var = 2;
        System.out.println("out="+var);
    }
    System.out.println("in="+var);
}

Does shadow the externally declared var . And it does output

in=1
out=2
in=1

Answer 2

Here you declare a variable and initialize it with value int var = 1;

Now you change it's value var = 2; , so value of var is 2.

System.out.println("in="+var); will print latest value of var which is 2.