CODEWITHSUNDEEP
javashadowing
Andrei Ciobanu
Andrei Ciobanu

Reputation: 12858

The concept of shadowing

Given the following code:

public class A {
 static final long tooth = 1L;

 static long tooth(long tooth){
  System.out.println(++tooth);
  return ++tooth;
 }

 public static void main(String args[]){
  System.out.println(tooth);
  final long tooth = 2L;
  new A().tooth(tooth);
  System.out.println(tooth);
 }
}

Can you please explain me the concept of shadowing ? And another thing, what tooth is actually used in the code from the main method ?

And i know it's a very ugly code, but ugly is the standard choice for SCJP book writers.

Upvotes: 2

Views: 763

Answers (2)

Alberto Zaccagni
Alberto Zaccagni

Reputation: 31580

When you are at this point

System.out.println(tooth);

the class property (static final long tooth = 1L;) is used, then a new tooth is declared, which shadows the class property, meaning that it is used instead of that.

Inside the tooth method the tooth variabile is passed as value, it will not be modified, you can see this by executing the main which gives:

1
3
2

Upvotes: 1

Gian
Gian

Reputation: 13955

There's nothing magical about shadowing as a concept. It's simply that a reference to a name will always be referencing the instance within the nearest enclosing scope. In your example:

public class A {
 static final long tooth#1 = 1L;

 static long tooth#2(long tooth#3){
  System.out.println(++tooth#3);
  return ++tooth#3;
 }

 public static void main(String args[]){
  System.out.println(tooth#1);
  final long tooth#4 = 2L;
  new A().tooth#2(tooth#4);
  System.out.println(tooth#4);
}

}

I've annotated each instance with a number, in the form "tooth#N". Basically any introduction of a name that is already defined somewhere else will eclipse the earlier definition for the rest of that scope.

Upvotes: 2

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