K&B chapter 8 Self Test Q 8

[Niranjan Deshpande]

public class Foo {
Foo() {System.out.print("foo");}

class Bar{
Bar() {System.out.print("bar");}
public void go() {System.out.print("hi");}
}

public static void main (String [] args) {
Foo f = new Foo();
f.makeBar();
}
void makeBar() {
(new Bar() {}).go();
}
}
//what is the result?
/*what is the result?
 *A. Compilation fails.
 *B. An error occurs at runtime.
 *C. foobarhi
 *D. barhi
 *E. hi
 *
 *line 8 seems to be making an anonymous instance as
 *new Bar() is followed by { }. Thus a ; must follow the second
 *bracket. Thus code should not compile
 *
 *But it does and answer is option C.
 *
 *Why?
 */

[ November 30, 2005: Message edited by: Barry Gaunt ]

[Ramakrishna RangaRaoOne AllamOne]

*line 8 seems to be making an anonymous instance as *new Bar() is followed by { }. Thus a ; must follow the second *bracket. Thus code should not compile
-----------------------------------------------


its an anonymous class , but u are not required to end an anonymous class with a semicolumn . the reason why u might thoght lik ethat is : it is afterall like a statement . But here the statement is anyways there . U are endng that statement with a ; . So must be ok ..

[james edwin]

Answer C is correct .Foo constructtor will be called first,then Bar constructor and then makeBar Method

[Gyanesh Sharma]

The assumption that the definition of an anonymous class must end with ; is correct only if the statement was an asignment statement. However in your code sample, an instance of the anonymous class is created and one of the method of the class is called in the same statement. So the ; goes in the end.
For example
a = new X();
a.doStuff();
can be written as new X().doStuff; but not as new X();.doStuff;

[Barry Gaunt]

The class Foo compiles and runs and prints out "foobarhi". So any assumption about a colon being necessary after the new Bar{}() is incorrect. That's it. It just creates a non-referenced object of an anonymous subclass of Bar. The code just happens to call the method doStuff(), inherited from Bar, for the side-effect of printing "hi". That's where the semicolon belongs, after the method call, and there it is. Because the object has no reference to it, it is immediately eligible for garbage collection.
[ December 01, 2005: Message edited by: Barry Gaunt ]

[rajendra naidu kumar]

DEAR RANCHERS,
PLZ HELP IAM PREPARING FOR A SCJP1.4;
IS IT SUFFICIENT K&B BOOK.

[Leoo yu]

public class Foo {
    Foo() {
        System.out.print("foo");
    }
 
    class Bar {
        Bar() {
            System.out.print("bar");
        }
 
        public void go() {
            System.out.print("hi");
        }
    }
 
    public static void main(String[] args) {
        Foo f = new Foo();
        f.makeBar();
    }
 
    void makeBar() {
        (new Bar() {
            public void go() {
                System.out.print("HIT");
        }}).go();
    }
} 

Result : foobarHIT
I hope this can help you .
Leoo yu

[Niranjan Deshpande]

I agree with you all....
but my question was what is the need to put { } after
Bar() ???

(new Bar() {}).go();

i know that

new Bar().go()

is as good as

Bar b=new Bar();
b.go()

but we never use (new Bar(){}).go( );
If you remove the {} the code compiles and runs fine

So the {} make it an unreferenecef anonymous class ???
then, whats the difference between
new Bar().go(); and
(new Bar() {}).go();

please help out
thanks in advance

[Jan Valenta]

Originally posted by Niranjan Deshpande:

So the {} make it an unreferenecef anonymous class ???
then, whats the difference between
new Bar().go(); and
(new Bar() {}).go();

try:

 (new Bar(){public void go(){System.out.println("hello");}}).go(); 

this is actualy overriding the go() method. So "(new Bar() {}).go();" is probably extending Bar but not making any overriding.