Static Members in a Class/Interface

[dmjpro]

What would be the better, whether should i access the static members using class name or using Object? Which one is preferred(low-cost effective)?

Today a thing stroke into my mind... Java supports multiple Interface Implementation but only single class extension to avoid ambiguous reference(data/members). Java manages to avoid method ambiguity but can't data ambiguity ;) Then the question stroke my mind static members should use using class name. But i am not sure which one is low-cost effective?

public class Test {
    int a = 100;
}

interface InterfaceTest{
    int a = 100;
}

class SubTest extends Test implements InterfaceTest{
    SubTest(){
        System.out.println("A: " + a); //Wrong one (ambiguous reference)
        System.out.println("A: " + InterfaceTest.a); //Right one
    }
}

[r035198x]

Use the class name always. Anyone looking at the code will be able to tell it's a static just by how you called it.

[dmjpro]

Use the class name always. Anyone looking at the code will be able to tell it's a static just by how you called it.

This is what you telling about coding convention.
My question was ..which one was cost-effective?

[JosAH]

This is what you telling about coding convention.
My question was ..which one was cost-effective?

I think you meant to write your example as:

public class Test { 
    static int a = 100; 
}

... and as said before always write Test.a, not o.a where o is an object of class Test. Also it makes more clear which static member you want. Compare the following (obfuscated) code snippet:

class B {
	public static int a= 42;
}

class D extends B {
	public static int a= 54;
}

public class Test {
	public static void main(String[] args) {

		B x= new D();
		
		System.out.println(x.a);
    }
}

What will be printed? 42 or 54? If you had written B.a or D.a it would've been clear immediately. There are no performance issues, better spend your energy on useful stuff (check it out with javap).

kind regards,

Jos

[dmjpro]

What will be printed? 42 or 54? If you had written B.a or D.a it would've been clear immediately. There are no performance issues, better spend your energy on useful stuff (check it out with javap).

Obviously the print will be 42. So you are telling that there is no performance issue..right! For coding convention it's better to use class name.

[Markus]

Sorry to jump in.

Jos, can you explain to me why your last code-snippet would return 42.

Thanks.

[JosAH]

Sorry to jump in.

Jos, can you explain to me why your last code-snippet would return 42.

Thanks.

Local variable x is of type B (I declared it that way so that's all the compiler knows) so x.a is B's static member variable a which is 42.

kind regards,

Jos

[Markus]

Local variable x is of type B (I declared it that way so that's all the compiler knows) so x.a is B's static member variable a which is 42.

kind regards,

Jos

So, if you create a new object of x, where x extends y, with the type of y, the new object will take on the members/properties of it's type (y) , not the class it created an object from (x)?

[JosAH]

So, if you create a new object of x, where x extends y, with the type of y, the new object will take on the members/properties of it's type (y) , not the class it created an object from (x)?

Yep, and the same counts for non-static members; remove the 'static' keyword in both classes in my example and you get 42 as output as well. The compiler decides which variable to use and from the 'outside world' the compiler knows no better than that x is a B; data members don't override they can only be hidden in the 'inside world' i.e. in code inside a method of a class.

Not so with method members: they are overridden ('virtual' functions in C++). That's why public data members are a no-no in object oriented programming. But, if the design of the language would've been such that data members were also overridden, the vft (Virtual Function Table) needs to contain pointers to all data members of (an object of) a class as well.

kind regards,

Jos

[Markus]

Yep, and the same counts for non-static members; remove the 'static' keyword in both classes in my example and you get 42 as output as well. The compiler decides which variable to use and from the 'outside world' the compiler knows no better than that x is a B; data members don't override they can only be hidden in the 'inside world' i.e. in code inside a method of a class.

Not so with method members: they are overridden ('virtual' functions in C++). That's why public data members are a no-no in object oriented programming. But, if the design of the language would've been such that data members were also overridden, the vft (Virtual Function Table) needs to contain pointers to all data members of (an object of) a class as well.

kind regards,

Jos

Data members are overridden, but member methods aren't. Understood.

Thanks, Jos.

[JosAH]

Data members are overridden, but member methods aren't. Understood.

Thanks, Jos.

Nonono, the terminology is: data members can be hidden and method members can be overridden; the first is a compiler issue, the second one is handled during runtime.

kind regards,

Jos

[dmjpro]

Obviously the print will be 42. So you are telling that there is no performance issue..right! For coding convention it's better to use class name.

Here is an another example ;)

public class Test {
    public static void main(String a[]){
        Derived d = new Derived();
    }
}

class Super{
    static int a = 100;
}

class Derived extends Super{
    static int a = 200;
    Derived(){
        System.out.println(super.a);
    }
}

So confusing ..but i have a warning "accessing static field".

[dmjpro]

Data members are overridden, but member methods aren't. Understood.

Thanks, Jos.

One more thing static members(only methods, as said my Josh ;) ) don't get overridden.

[dmjpro]

Nonono, the terminology is: data members can be hidden and method members can be overridden; the first is a compiler issue, the second one is handled during runtime.

kind regards,

Jos

Have a look at this example @Markus

class Super{
    int b = 100;
}

class Derived extends Super{
    int b = 200;
}

//Main Program.
public static void main(String a[]){
        Super d = new Derived();
        System.out.println("The value of b: " + d.b);
    }

[JosAH]

Have a look at this example @Markus

Don't make this an obfuscated thread; we were discussing static members. Don't throw in a completely other example or add a bit of explanatory text or a clear question.

kind regards,

Jos