Problems with super ...

[Emilian Chmiel]

Hi, I wrote simple a program:

class first
{
	int i;
}
class second extends first
{
	second (int a, int b)
		{
		super.i=a; i=b;
		}
	void print()
		{
		System.out.println("i "+i+"; super.i: "+super.i);
		}
}
public class UseSuper 
{

	public static void main(String[] args) 
	{
		second test = new second(1, 2);
		test.print();
	}

}

when I run i I get:

i 2; super.i: 2

should't I get

i 1; super.i: 2

[Henry Wong]

Line 9 is merely two different ways to access the same variable. You don't have two variables -- as there is only one declaration.

Henry

[Ganesh Patekar]

Emilian Chmiel wrote:when I run i I get:

i 2; super.i: 2

should't I get

i 1; super.i: 2

No, If you see here

second (int a, int b){
    super.i=a; i=b;
}

The formal parameters int a has value 1 and int b has 2 value passed by this code

second test = new second(1, 2);
// invokes parameterized constructor of second and assign 1 to a and 2 to b

then in that constructor of second we have

super.i = a;

Which assigns value of a i.e. 1 to instance variable i of super class first. Now we have i = 1

super is the keyword used to access field of super class, here instance variable int i of super class first.

But here on the same line

i=b;

We again assigned i = b where value of b is 2. Now value of i is 2.

After above assignments, we called to print() method which prints

System.out.println("i "+i+"; super.i: "+super.i);

So i and super.i both means same variable of class first which has value 2 so you get output i 2; super.i: 2

[Ganesh Patekar]

Name of class always starts with capital letter as you named UseSuper, so becomes easy to identify it. Worth reading Java Programming Style Guide