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8: Inner Classes
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AbstractTest.Bar f = t.new Bar() { public int getNum() { return 57; } }; System.out.println(f.getNum() + " " + t.getNum()); } }
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A. 57 22 B. 45 38 C. 45 57 D. An exception occurs at runtime. E. Compilation fails.
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1. A. MyInner is a static nested class, so it must be instantiated using the fully-scoped name of MyOuter.MyInner. B is incorrect because it doesn t use the enclosing name in the new. C is incorrect because it uses incorrect syntax. When you instantiate a nested class by invoking new on an instance of the enclosing class, you do not use the enclosing name. The difference between A and C is that C is calling new on an instance of the enclosing class rather than just new by itself. D is incorrect because it doesn t use the enclosing class name in the variable declaration. 2. B and D. B is correct because a static nested class is not tied to an instance of the enclosing class, and thus can t access the nonstatic members of the class (just as a static method can t access nonstatic members of a class). D uses the correct syntax for instantiating a static nested class. A is incorrect because static nested classes do not need (and can t use) a reference to an instance of the enclosing class. C is incorrect because static nested classes can declare and define nonstatic members. E is wrong because it just is. There s no rule that says an inner or nested class has to extend anything. 3. E is correct. It defines an anonymous inner class instance, which also means it creates an instance of that new anonymous class at the same time. The anonymous class is an implementer of the Runnable interface, so it must override the run() method of Runnable. A is incorrect because it doesn t override the run() method, so it violates the rules of interface implementation. B, C, and D use incorrect syntax. 4. B and C. B is correct because anonymous inner classes are no different from any other class when it comes to polymorphism. That means you are always allowed to declare a reference variable of the superclass type and have that reference variable refer to an instance of a subclass type, which in this case is an anonymous subclass of Bar. Since Bar is a subclass of Boo, it all works. C uses correct syntax for creating an instance of Boo. A is incorrect because it passes an int to the Boo constructor, and there is no matching constructor in the Boo class. D is incorrect because it violates the rules of polymorphism you cannot refer to a superclass type using a reference variable declared as the subclass type. The superclass is not guaranteed to have everything the subclass has. E uses incorrect syntax.
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5. B is correct because the syntax is correct using both names (the enclosing class and the inner class) in the reference declaration, then using a reference to the enclosing class to invoke new on the inner class. A, C, D, and E all use incorrect syntax. A is incorrect because it doesn t use a reference to the enclosing class, and also because it includes both names in the new. C is incorrect because it doesn t use the enclosing class name in the reference variable declaration, and because the new syntax is wrong. D is incorrect because it doesn t use the enclosing class name in the reference variable declaration. E is incorrect because the new syntax is wrong. 6. B and E. B is correct because a method-local inner class can be abstract, although it means a subclass of the inner class must be created if the abstract class is to be used (so an abstract method-local inner class is probably not useful). E is correct because a method-local inner class works like any other inner class it has a special relationship to an instance of the enclosing class, thus it can access all members of the enclosing class. A is incorrect because a method-local inner class does not have to be declared final (although it is legal to do so). C and D are incorrect because a method-local inner class cannot be made public (remember you cannot mark any local variables as public), or static. 7. C is correct because the syntax of an anonymous inner class allows for only one named type after the new, and that type must be either a single interface (in which case the anonymous class implements that one interface) or a single class (in which case the anonymous class extends that one class). A, B, D, and E are all incorrect because they don t follow the syntax rules described in the response for answer C. 8. C is correct because first the Foo instance is created, which means the Foo constructor runs and prints foo . Next, the makeBar() method is invoked which creates a Bar, which means the Bar constructor runs and prints bar , and finally the go() method is invoked on the new Bar instance, which means the go() method prints hi . A, C, D, E, and F are incorrect based on the program logic described above. 9. G. This code would be legal if line 7 ended with a semicolon. Remember that line 3 is a statement that doesn t end until line 7, and a statement needs a closing semicolon! A, B, C, D, E, and F are incorrect based on the program logic described above. If the semicolon were added at line 7, then answer B would be correct the program would print true , the return from the equals() method overridden by the anonymous subclass of Object.
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