Question

So we all know that all classes implicitly extend Object. How about interfaces? Is there an implicit super-interface? I say there is. The following code compiles:

java.io.Serializable s1 = null;
java.io.Serializable s2 = null;

s1.equals(s2);

The equals method is not declared in Serializable, but in Object. Since interfaces can only extend other interfaces, and Object is a class, not an interface, there must be some implicit interface that is being extended. And the Object class must then implicitly implement this implicit interface (wow, that was weird to write).

So, the question is, how correct is this?

Answer 1

Since interfaces can only extend other
interfaces, and Object is a class, not
an interface, there must be some
implicit interface that is being
extended.

No. Citing the Java Language Specification:

If an interface has no direct
superinterfaces, then the interface
implicitly declares a public abstract
member method m with signature s,
return type r, and throws clause t
corresponding to each public instance
method m with signature s, return type
r, and throws clause t declared in
Object, unless a method with the same
signature, same return type, and a
compatible throws clause is explicitly
declared by the interface. It is a compile-time error if the interface explicitly
declares such a method m in the case
where m is declared to be final in
Object.

The difference between this and your “implicit super interface” is that Object has a number of final and protected methods, and you couldn’t have those modifiers in an interface.