When you define a new interface, you are defining a new reference data type. You can use interface names anywhere you can use any other data type name. If you define a reference variable whose type is an interface, any object you assign to it must be an instance of a class that implements the interface.As an example, here is a method for finding the largest object in a pair of objects, for any objects that are instantiated from a class that implements
Relatable:By castingpublic Object findLargest(Object object1, Object object2) { Relatable obj1 = (Relatable)object1; Relatable obj2 = (Relatable)object2; if ( (obj1).isLargerThan(obj2) > 0) return object1; else return object2; }object1to aRelatabletype, it can invoke theisLargerThanmethod.If you make a point of implementing
Relatablein a wide variety of classes, the objects instantiated from any of those classes can be compared with thefindLargest()method—provided that both objects are of the same class. Similarly, they can all be compared with the following methods:These methods work for any "relatable" objects, no matter what their class inheritance is. When they implementpublic Object findSmallest(Object object1, Object object2) { Relatable obj1 = (Relatable)object1; Relatable obj2 = (Relatable)object2; if ( (obj1).isLargerThan(obj2) < 0) return object1; else return object2; } public boolean isEqual(Object object1, Object object2) { Relatable obj1 = (Relatable)object1; Relatable obj2 = (Relatable)object2; if ( (obj1).isLargerThan(obj2) == 0) return true; else return false; }Relatable, they can be of both their own class (or superclass) type and aRelatabletype. This gives them some of the advantages of multiple inheritance, where they can have behavior from both a superclass and an interface.
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