doubt in shadowing

[dolly shah]

public class knowing {
    static final long tooth=343L;
    static long doit(long tooth){
        System.out.println(++tooth + "");
        return ++tooth;
    }
    public static void main(String... args){
        System.out.println(tooth + "");
        final long tooth=340L;
        new knowing().doit(tooth);
        System.out.println(tooth);
    }
}

Answer:343 341 340

This is from chapter:3, example:9, page no:270 from K & B book.

Book describe " There are three different long variables named tooth. Remember that you can apply the final modifier to local variables, but in this case the 2 versions of tooth marked final are not changed. The only tooth whose value changes is the one not marked final. This program demonstrates a bad practice known as shadowing."

I think both instance & local variable are final. You cannot change any of them. Book says "you can change which is not marked final".
I am not getting which variable is not final?

Can anyone please explain?

[ahmed yehia]

Here the variable that is not final is the variable in the method parameter as in: static long doit(long tooth){

[dolly shah]

But from the main method code passes it "tooth" variable which is final. How can you change it? Suppose if you change it then its value will be "342". if I am wrong ,correct me.

[Remko Strating]

You get a copy of the variable tooth because variable tooth is a primitive. The copy of the variable tooth lives only within the method doit and is not declared final.

[dolly shah]

I got it. Thanks.