how to define a variable of a class type in java when the class has extends?

Question

i have this class:

   public class FindPath<T,N extends Path<T,N>> {

    public FindPath() {
        
   // the constructor ...
    }
}

and the path interface:

public interface Path<N, P extends Path<N,P>>
            extends Iterable<N>, Comparable<Path<?,?>>

where T, P and N are generic types. i want to define a variable FindPath in another class, and call it's constructor, basically i want something like this:

public class TestD {

private FindPath <WNode,WNodePath extends Path<WNode,WNodePath>> Find_Path = new FindPath;

} 

( WNode,WNodePath are another classes i use in the generic types )

what am i doing wrong? how to do this right?

also another question:|| the WNodePath class is defined like so:

public class WNodePath implements Path<WNode, WNodePath > { .. }

the path interface is Comparable , does that mean that also WNodePath is Comparable, which means that i can use it compareTo() function in a priority queue?

Answer 1

When you declare a FindPath reference you don't use extends , that's used to provide bounds (limits) on the classes it is legal to use for that type parameter. (You can use extends when you have a wildcard type parameter, but that's not relevant here, see https://docs.oracle.com/javase/tutorial/extra/generics/wildcards.html )

Just say:

    private FindPath<WNode,WNodePath> findPath = new FindPath<>();

Which compiles.

Yes, WNodePath implements Comparable , transitively via Path .