super instanceof Class?

[mikeiz404]

Is there any way to determine if the parent class of an object is an instance of another class?

This seems not to work, but I am unsure why:
if(super instanceof Object){}

[angryboy]

keyword: instanceof

[mikeiz404]

that doesn't seem to work with the super keyword

[mikeiz404]

I've seemed to have found a work around which uses a helper class.
If anyone one knows the answer to the original question, I'd be interested to know how. Thanks.

[Fubarable]

It kind of begs the question: why do you want to do this? It smells of design problems.

[mikeiz404]

I'm trying to create a series of filters for data. For example (types of data filters and their hierarchy):
.................[TCP]..................
........[Web]..........[E-mail]......
[images]..[html]..[pop]...[smtp]

So it seemed logical to create an Abstract filter class:

public abstract class Filter2{
	public final boolean isTrue(Packet p){
		if(super instanceof Filter2){ //and super is not equal to filter
			return evaluate(p) && super.isTrue(p);
		}
	}	
	public abstract boolean evaluate(Packet p);
	
	public final int getFilterLevel(){
		if(super instanceof Filter2){ //and super is not equal to filter
			return 1+super.getFilterLevel();
		}
	}
	
	public final LinkedList<Class> getFilterLevels(){ // not sure how useful this method really is
		if(super instanceof Filter2){ //and super is not equal to filter
			LinkedList<Class> l = super.getFilterLevels();
			l.add(super.getClass());
			return l;
		}else{
			return new LinkedList<Class>();
		}
	}
}

This way it would be much easier to create a TreeMap of the filters and sort the tree accordingly.
Is there a better/easier way to do this?

[angryboy]

=\ no idea what your code does. but you can use the class Class. <= no thats not a typo.

if(this.getClass().getSuperclass().getSimpleName().equals("Filter2")) ...

you can put that in a method to make it easier to read.

[DarrylBurke]

Shorter:

if(this.getClass().getSuperclass() == Filter2.class) ...

db

[Steve11235]

I see what you are up to. This is not a good approach.

If you want to determine the type of an object, just use something like (assuming Filter2 is the base of your hierarchy)

public String getClassType(final Filter2 pInstance) {
  final String className;
  className = pInstance.getClass.getName();
  return className;
}

This will return the fully qualified name of the class, like java.lang.String. You can then use String.startsWith() to see if it is part of a particular hierarchy.

[angryboy]

Shorter:

if(this.getClass().getSuperclass() == Filter2.class) ...

db

where in the jdk doc can I find more info on X.class ? Its not listed under java.lang.Object. what class is this field inherited from?

Any other .stuff ?? (.length, .class, what else?)

[Steve11235]

The main reason to use "super" is to access overridden methods of the base class.

If "super instanceof SomeClass" is true, then "this instanceof SomeClass" must also be true. The same holds for false. The only exception is if we replace SomeClass with SomeInterface. This discussion is part of the reason Java only allows single inheritance.

There is no need to resort to "super".

Be careful of getClass(). The only time super.getClass() == Filter2.class() is true is if this directly extends Filter2. If this extends That extends Filter2, then the comparison will fail, since super.getClass will return That, not Filter2. instanceOf goes through the entire hierarchy.

[mikeiz404]

Thanks for all your replies, they have been helpful. If anyone is interested, here is the code I have come up with which seems to work:

	public LinkedList<Class> getFilterLevels(){
		Class parent = this.getClass();
		LinkedList<Class> l = new LinkedList<Class>();
		while(!parent.equals(Filter.class)){
			l.add(parent);
			parent = parent.getSuperclass();
		}
		return l;
	}