Output of Java program

Question

I am a Java beginner. Can anyone explain why is it printing output 2?

interface Foo { int bar(); } public class Beta { class A implements Foo { public int bar() { return 1; } } public int fubar(final Foo foo) { return foo.bar(); } public void testFoo()// 2 { class A implements Foo { public int bar() { return 2; } } System.out.println(fubar(new A())); } public static void main(String[] args) { new Beta().testFoo(); } }

Naftali · Accepted Answer · 2011-05-03 15:21:57Z

5

That is because you redefined Class A here:

 class A implements Foo { public int bar() { return 2; } } System.out.println(fubar(new A()));

So when you do return foo.bar(); you return 2

answered May 3, 2011 at 15:21

Naftali

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4 Comments

Adrian Mouat Over a year ago

I think the confusion may be caused because the call to "new A()" is happening within the context of the testFoo() method, not fubar().

Dilum Ranatunga Over a year ago

+1 for speedy correct response. But @Neal, I like to avoid the term 'redefine' because in some other languages, you can actually redefine the method itself. I like the term 'hide'. Thoughts?

Naftali Over a year ago

@Dilum, yes, but how do you put hide in this context?

Dilum Ranatunga Over a year ago

I guess something along the lines of 'within Beta.testFoo(), the inner class Beta.A is hidden by the method-scoped inner class A, because the name is re-used.'

Community · Accepted Answer · 2017-05-23 12:19:00Z

Because the innermost definition of A is in the testFoo() method, and its method bar() return 2.

You may also find the answer to my question from today interesting.

Kal · Accepted Answer · 2011-05-03 15:24:06Z

When you say, System.out.println(fubar(new A()));

the class A created is the one defined inside testFoo().

Dilum Ranatunga · Accepted Answer · 2011-05-03 15:26:56Z

There are many places in java where you can hide a broader name with a more local name. This is true of parameters vs member variables, class names etc. In your case, you are hiding Beta.A with the A you defined in the method.

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