Blog

With the Super keyword, you can:

Look at the following example:

public class Test {
	
	public static void main(String[] args)
	{
		Square mysquare = new Square();
		mysquare.getInformation();
		
	}
}


class Shape // the SUPERCLASS
{
	int color = 4;
			
	public void getInformation() // getInformation() in the superclass
	{
		System.out.println(color);
	}
}

class Square extends Shape // the SUBCLASS
{
	String sides="4 sides";
	
	public void getInformation() // getInformation() in the subclass
	{
		getInformation();
	}
}

This outputs:

Exception in thread “main” java.lang.StackOverflowError
at Square.getInformation(Test.java:30)
at Square.getInformation(Test.java:30)
at Square.getInformation(Test.java:30)

Why the error?

At line 28, the call to method getInformation() actually has “this” appending to it automatically:

this.getInformation()

“this” refers to the object that called it “mysquare”

So what happens is a cycle that goes round and round and crashes the program.

However, if we use the “super” keyword

super.getInformation() as follows on line 28

public class Test {
	
	public static void main(String[] args)
	{
		Square mysquare = new Square();
		mysquare.getInformation();
		
	}
}


class Shape // the SUPERCLASS
{
	int color = 4;
			
	public void getInformation() // getInformation() in the superclass
	{
		System.out.println(color);
	}
}

class Square extends Shape // the SUBCLASS
{
	String sides="4 sides";
	
	public void getInformation() // getInformation() in the subclass
	{
		super.getInformation();
	}
}

This outputs:

4

The getInformation() method in the SUPERCLASS is being called (the Shape class)