Static class and Dynamic class when using a constructor

Question

Say that I have a super class and a sub class like below.

class Super { 
 public void action( ) { 
 System.out.println( “Super’s action” ); 
 } 
} 

class Sub extends Super( ) { 
 public void action( ) { 
 System.out.println( “Sub’s action” ); 
 } 
}

In the main method I call this.

Super s1 = new Super( ); 
s1.action( ); 

Super s2 = new Sub( ); 
s2.action( ); 

Sub s3 = new Sub( ); 
s3.action( );

I know that the output will be

Super’s action 
 Sub’s action 
 Sub’s action

I've learnt that in the constructor, the first word is the static class type and the part with the () is the dynamic class. I know that the method lookup will always begin in the dynamic class so what I want to know is, what is the difference between s2 and s3? i.e Having the static class be of the parent type, or the static class be it's own type

Put another method in `Sub`, try to call it from `s2` and `s3`, and you'll answer your own question. — awksp, Jun 16 '14 at 23:14
In what context? They are simply declared with a different type, their compile time (or static) type. — Sotirios Delimanolis, Jun 16 '14 at 23:14
Take a look at my answer [here](http://stackoverflow.com/a/24113813/391161). — merlin2011, Jun 16 '14 at 23:27
Thanks ajb and merlin2011. That answered my question perfectly — Jared Y, Jun 16 '14 at 23:38

score 0 · Answer 1 · answered Jun 16 '14 at 23:15

0

If Sub defines a method that isn't in Super, then you can't call it from s2. Also, if a method takes an argument of type Sub, you can't pass s2 to it.

answered Jun 16 '14 at 23:15

Ismail Badawi

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Static class and Dynamic class when using a constructor

1 Answers1