Output of the Program

[Kalai Selvi]

The Piece of code goes like this:

public class A
{
A()
{
}
}
1.The class A can be referenced outside the package in which it is defined.
2.The class A cannot be instantiated outside the package in which it is defined.
3.The class A cannot be extended outside the package in which it is defined.
4.The class A can be referenced, instantiated or extended anywhere.
5,The above code will cause a compiler error. The constructors of public class have to be public.

[m Sabale]

the answer should be 4.
since class A is Public, JVM will implicitly assign public access modifier to constructor of A.

[Kaydell Leavitt]

Isn't the answer 2 since having no access qualifier is the default which is package access?

Kaydell

[Tannis Bo]

I would've gone with 4 as well...

[Cao Bin]

I have just made a test.

when compile,the compile error like this:
A() in a.A is not public.so can't be get from other packages.

so 2,3 should be choosed

[krishna bulusu]

It can not be instantiated as well as extended outside the package as constructor of A is not visible over there.

[Ankith suresh]

My choice is 4
Since class is public constructor should be public. right?

[Manfred Klug]

Since the class is public, it can be referenced outside of the package.

So 1, 2 and 3 are correct.

[anil kumar]

Hi

suresh

NO.

Because it is not provided by the compiler.It is provided by the programmer.If it is provided by the compiler then the accessmodifier it will get is the modifier specified in the class.




Thanks

Anil Kumar

[Ankith suresh]

Hi anil

You are right. Thank you.
Then what is the answer here

[Ankith suresh]

package pk1;
public class SCJP11
{
SCJP11()
{
}
}
---------------------------------

package pk2;
import pk1.SCJP11;
class SCJP11Test extends SCJP11 //1
{

}

When i compile the SCJP11Test.java
i am getting following error

SCJP11Test.java:3: SCJP11() is not public in pk1.SCJP11; cannot be accessed from outside package
class SCJP11Test extends SCJP11
^
1 error

Why the error is displaying while extending..?
Compiler trying to access the constructor while extending the class?
I thought error will display only when trying to create the object.

[Raghavan Muthu]

Hi,

I have also tested now. I use JDK 1.4.2_12 (fyi and if reqd). I have not got any exceptions or errors. The code compiled fine.

    public class A 
    { 
      A() 
      { 
      } 
    }
   1.The class A can be referenced outside the package in which it is defined. 
   2.The class A cannot be instantiated outside the package in which it is defined. 
   3.The class A cannot be extended outside the package in which it is defined. 
   4.The class A can be referenced, instantiated or extended anywhere. 
   5.The above code will cause a compiler error. The constructors of
public class have to be public. 
  

Lets look at the options one by one....

option 1:
It holds good as the access modifier of class is given as public. So its valid and can be chosen.

option 2:
It holds good because the class is of default package and the constructor is not public!

option 3:
I think this will be ruled out! because, extension does check the class's access level and the class is of public access, it should be allowed to be extended.

option 4:
It is not applicable because of action 2 is ruled out (instantation).

option 5:
It does not hold good as there are no hard and fast rules of the public class to have the constructor as public. It can have non-public constructors as well. (thats why the above program compiled fine without any exceptions)

HtH.
[ June 12, 2007: Message edited by: Raghavan Muthu ]

[krishna bulusu]

4.The class A can be referenced, instantiated or extended anywhere.
see the code below:

package p1;
 
class A {
A(){

}
}
public class B extends A{
public static void main(String[] args) {
A a = new A();
}
}
 

According to above code, the option 4 is ruled out.
The answere should be 2,3.

[anil kumar]

Hi

suresh


The answers are 2 and 3


Thanks

Anil Kumar

[Ishita Saha]

Hi,

I tried creating a reference of class A in another class B which is declared in different package.

So i just said A a1;
and it works.
the compiler complains only when i say:

a1 = new A();
so going by that 1,2,3 should be the answer.

[anil kumar]

Hi

Saha
--------------------------------------------------------------
So i just said A a1;
----------------------------------------------------------------

Ya what you said is true.But here a1 is not refering to any thing.You just declare a1 of type A.If it tries to refer to A or any of its subclass ,you will get error.


Thanks

Anil Kumar

[Ishita Saha]

but then what would be the difference between instantiating A and creating a reference to A i.e option 1 & 2

[krishna bulusu]

but then what would be the difference between instantiating A and creating a reference to A i.e option 1 & 2

Instantiating means creating the object.
I am instatiating the class A means, i am making an object of class A.
A a = new A();
I am creating a reference of A means, simple declaring a variable as type A.
A a;
Here a is the identifier of type A.
you can assign an object of type A to a or the object of its child class.
Hope you got the point.

[anil kumar]

Hi
--------------------------------------------------------------------------
1.The class A can be referenced outside the package in which it is defined.
2.The class A cannot be instantiated outside the package in which it is defined.
---------------------------------------------------------------------------
1) false

Here A can be referenced, means we can assign the object of type A or sub type of A,which we can't do because the sub clsss can't find the constructor of the super class.
2)true
means A a1=new A();

we can't do this because A is in the other pacakge.




Thanks

Anil Kumar

[Raghavan Muthu]

I think what Anil said is right.

Ishita Saha,

Hope krishna answered your question. To summarize,

     A a; //declare a reference variable
     a = new A();//instantiating class A
   

Krishna,

I think the sample code you had given to explain is having both the classes A and B in one package. But in the question given by OP does not mention about anything and its an abstract term when given in the options.

HtH.

[krishna bulusu]

Anywhere means every place like same package, different package etc etc.
but, we can do it in the same package level.
so, every where is ruled out. it should be some where means to a particular area.

[Raghavan Muthu]

Krishna,

I think the term "anywhere" in the question is quite subjective and with respect to this scenario, i tend to interpret it as "everywhere" only. Means, in all the places without any exception.

Am i right? Or is it the other way around as you said?

Ranchers, please clarify!

[Manfred Klug]

Hi,

a small sample that demonstrates that option 1 is correct.

package package1;
 
public class A {
    A() {}
    
    public void doIt() {
        System.out.println("A.doIt()");
    }
 
    public static A getA() {
        return new A();
    }
}

package package1;
 
public class SubA extends A {
    public SubA() {
        super();
    }
}

package package2;
 
import package1.A;
import package1.SubA;
 
public class Test {
 
    public static void main(String[] args) {
        A a1 = A.getA();
        A a2 = new SubA();
 
        a1.doIt();
        a2.doIt();
    }
 
}

[Raghavan Muthu]

Hi Manfred,

Thats fine with your code. But the OP has posted only with the constructor with default visibility. Thats where the question started off.

[Manfred Klug]

Originally posted by Raghavan Muthu:

But the OP has posted only with the constructor with default visibility. Thats where the question started off.

Even then can A be referenced outside the package.

package package1;
 
public class A {
    A() {}
}

package package1;
 
public class SubA extends A {
    public SubA() {
        super();
    }
}

package package2;
 
import package1.A;
import package1.SubA;
 
public class Test {
    public static void main(String[] args) {
        A a2 = new SubA();
    }
}

[Sasha Ruehmkorf]

Just to be sure, that Ankith suresh's post is not being overlooked:

Option 3 is false. See post at 9:29 and to answer your question Ankith: If you don't give a constructor explicitly in the derived class, the compiler adds a default-constructor including a call to super() and this called constructor is not visible, so of course the compiler complains

1: yes
2: yes
3: no
4: no
5: no

[Edit]

And as there seems to be confusion around option 1, just see this code of three files:

package pp1;
 
public class A {
A(){}
}
 
____________________
 
package pp1;
 
public class GiveA {
public Test method(){
return new A();
}
}
 
____________________
 
package pp2
 
public class B{
A myReference = pp1.GiveA();
}
 
____________________

A is referenced in another package.

[Edit 2]

Sorry Manfred, you already made it clear
[ June 12, 2007: Message edited by: Sasha Ruehmkorf ]

[Manfred Klug]

Hi Sasha,

a small correction. Option 3 is true.

3.The class A cannot be extended outside the package in which it is defined.

[ June 13, 2007: Message edited by: Manfred Klug ]

[Sasha Ruehmkorf]

I'll argue the converse. How comes, I was sure that point 3 stated that it is possible to extend from outside the package.

My certification exam starts in 2 hours and it seems like I have to read the questions and answers more carefully

[Raghavan Muthu]

it seems like I have to read the questions and answers more carefully

Yes, thats very much true. I remember one of the ranchers was having a quote in his signature as, "We should know the questions properly before the answers!" (the sentence may not be appropriate)

Good luck!

[Thillakan Saba]

My choices are 2 and 3.

[Gaurav Pavan Kumar Jain]

HI

Answer must be 4

[raj malhotra]

public class A 
{ 
A() 
{ 
} 
}

1.The class A can be referenced outside the package in which it is defined.

class A can be referenced outside the package if we import the class in other package and its constructor is public. which is not the case so this is not the answer.

2.The class A cannot be instantiated outside the package in which it is defined.

Even if we import the class A,we cannot instantiate there as constructor is not visible.so This answer is ok

3.The class A cannot be extended outside the package in which it is defined.

class A cannot be extended because subclass constructor need to call super constructor which is not visible so class A cannot be extended outside the package.so this is also correct.

4.The class A can be referenced, instantiated or extended anywhere.

only in the same package.so this is not correct.

5,The above code will cause a compiler error. The constructors of public class have to be public.

we can specify any access modifier for a top level class.so this is also not correct
[ June 14, 2007: Message edited by: raj malhotra ]

[Sergio Tridente]

Originally posted by raj malhotra:
class A can be referenced outside the package if we import the class in other package and its constructor is public. which is not the case so this is not the answer.

Option one is OK. Class A can be referenced outside package, it does not matter if it cannot be intantiated. Take for instance

package scjp.first.pack;
 
public class A {
A() {}
}

package scjp.first.pack;
 
public class Creator {
public A getAInstance() {
return new A();
}
}

package scjp.second.pack;
 
import scjp.first.pack.*;
 
public class Referator {
public static void main(String... args) {
A myA = new Creator().getAInstance();
}
}

You can refer to A in second package, but you still cannot create an instance of A, so we used a creator class that 'lives' in the first package.
[ June 14, 2007: Message edited by: Sergio Tridente ]

[raj malhotra]

yes Sergio Tridente! you are correct .Thanks

[krishna bulusu]

1.The class A can be referenced outside the package in which it is defined.

yes, it seems the above statement is right.