58. You have these files in the same directory. What will happen when you attempt to compile and run Class1.java if you have not already compiled Base.java
//Base.java
package Base;
class Base{
protected void amethod(){
System.out.println("amethod");
}//End of amethod
}//End of class base
package Class1;
//Class1.java
public class Class1 extends Base{
public static void main(
String argv[]){
Base b = new Base();
b.amethod();
}//End of main
}//End of Class1
1) Compile Error: Methods in Base not found
2) Compile Error: Unable to access protected method in base class
3) Compilation followed by the output "amethod"
4)Compile error: Superclass Class1.Base of class Class1.Class1 not found
the ans is 4 but I chose 1. I don't understand what the compile error in 4 said about.
59.class Base{
private void amethod(int iBase){
System.out.println("Base.amethod");
}
}
class Over extends Base{
public static void main(String argv[]){
Over o = new Over();
int iBase=0;
o.amethod(iBase);
}
public void amethod(int iOver){
System.out.println("Over.amethod");
}
}
1) Compile time error complaining that Base.amethod is private
2) Runtime error complaining that Base.amethod is private
3) Output of "Base.amethod"
4) Output of "Over.amethod"
the ans is 4. In this overriding, I understand the name of parameter doesn't matter, I want to ask is the base class declare private but supclass declare public, why they still can override?
Kindly explain these for me!!
Thanks
