I would like to understand what this Java declaration is doing:
public static <AnyType extends Comparable<? super AnyType>> int mymethod( AnyType x ) {
/* ... */
}
From my basic knowledage of Java, I think all it's doing is telling you that the param x must be of any type, but that type must extend Comparable?
Your understanding is correct. This method indicates that it takes an argument of a type that extends a Comparable (Note that I would have called the type parameter T instead of AnyType for readability).
Now for the super in Comparable<? super AnyType>>, it means that this comparable implementation can actually be, for example, an implementation of Comparable<Object>, that is a comparable type that can be compared to an object. More generally, the type accepted by the method can be a Comparable which can be compared to some type that is a superclass or superinterface of it, hence the keyword super. In other words, the method can be invoked as follows:
// An object of this type can be compared to an Object
class X implements Comparable<Object> {
@Override
public int compareTo(Object o) {
...
}
}
X x = new X();
mymethod(x);
This is a generic method, i.e. its type parametrization is not related to the class' parametrization if any.
The <AnyType extends Comparable<? super AnyType>> part is the generic parametrization of the method.
The extends keyword should not be confused with the extends keyword when declaring a class: in this case it looks more like "is assignable from", in other words, IS-A.
Comparable<T> is an interface - see docs here.
Your method parametrization will make it so that your method accepts any parameter, as long as its type implements Comparable.
Specifically, it needs to be a Comparable of itself or any super-type (that's what the super part means, as opposed to extends).
For instance, you can pass a String argument because String implements Comparable<String> (amongst others).
