Access modifiers problem.

[sharma ishu]

Given the following method signatures from ArrayList:
boolean add(E e)
protected void removeRange(int fromIndexInclusive, int toIndexExclusive)

int size()and given:

import java.util.*;
public class MyUtil extends ArrayList{
public static void main(String[] s){
Myutil m=new MyUtil();
m.add("w"); m.add(""x); m.add("y"); m.add("z"); //#1
m.removeRange(1,3);
MyUtil m2=new MyUtil2().go();
}
}

class MyUtil2{
MyUtil go(){
MyUtil m2=new MyUtil();
m2.add("1"); m2.add("2"); m2.add("3"); //#2
m2.removeRange(1,2); //#3
return m2;
}
}

Shouldn't #1 and #3 cause compiler errors as add()method has default access and cannot be inherited? And also at #3, why does this cause compiler error?

This question is from Practic Exam book by Kethy and Bates 2011.(Page 88).

[fred rosenberger]

Please read our FAQ on how to UseCodeTags (<--that is a link). Unformatted code is hard to read. Many folks will see the unformatted source and decide it is not worth their valuable time to try and decipher what you wrote. Posting properly formatted code will most likely help you get an answer faster.

[Henry Wong]

sharma ishu wrote:Given the following method signatures from ArrayList:

boolean add(E e)
protected void removeRange(int fromIndexInclusive, int toIndexExclusive)

int size()

and given:

import java.util.*;
public class MyUtil extends ArrayList{
  public static void main(String[] s){
     Myutil m=new MyUtil();
     m.add("w"); m.add(""x); m.add("y"); m.add("z");                             //#1
     m.removeRange(1,3);
     MyUtil m2=new MyUtil2().go();
 }
}

class MyUtil2{
  MyUtil go(){
     MyUtil m2=new MyUtil();
     m2.add("1"); m2.add("2"); m2.add("3");                                           //#2
     m2.removeRange(1,2);                                                                  //#3
     return m2;
  }
}

Shouldn't #1 and #3 cause compiler errors as add()method has default access and cannot be inherited? And also at #3, why does this cause compiler error?

This question is from Practic Exam book by Kethy and Bates 2011.(Page 88).

The add() method of the ArrayList class is declared as public -- see JavaDoc ... http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html#add%28E%29

Henry

[sharma ishu]

So, you mean to say the question in the book is wrong?

[sharma ishu]

Henry Wong wrote:

sharma ishu wrote:Given the following method signatures from ArrayList:

boolean add(E e)
protected void removeRange(int fromIndexInclusive, int toIndexExclusive)

int size()

and given:

import java.util.*;
public class MyUtil extends ArrayList{
  public static void main(String[] s){
     Myutil m=new MyUtil();
     m.add("w"); m.add(""x); m.add("y"); m.add("z");                             //#1
     m.removeRange(1,3);
     MyUtil m2=new MyUtil2().go();
 }
}

class MyUtil2{
  MyUtil go(){
     MyUtil m2=new MyUtil();
     m2.add("1"); m2.add("2"); m2.add("3");                                           //#2
     m2.removeRange(1,2);                                                                  //#3
     return m2;
  }
}

Shouldn't #1 and #3 cause compiler errors as add()method has default access and cannot be inherited? And also at #3, why does this cause compiler error?

This question is from Practic Exam book by Kethy and Bates 2011.(Page 88).

The add() method of the ArrayList class is declared as public -- see JavaDoc ... http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html#add%28E%29

Henry

Please also explain me why does #3 cause compiler error?

[Henry Wong]

sharma ishu wrote:So, you mean to say the question in the book is wrong?

When in doubt, looking at the official documentation is a very good habit to get into.... in fact, the JavaDoc should always be a click away.

sharma ishu wrote:

Henry Wong wrote:
The add() method of the ArrayList class is declared as public -- see JavaDoc ... http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html#add%28E%29

Please also explain me why does #3 cause compiler error?

Well, what does the JavaDoc say?

Henry

[Matt Baston]

Thanks, this makes sense.