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classes in different package

 
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Hi
Pl have a look at the undergiven code.
The correct answer is 4.
Can anybody explain this error.

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
 
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I missed this one too. I don't think the explanation given makes any sense. How can you run the file if there's no public static void main( String arg[]) there?
See my point?
Regards,
Khalid
 
Anonymous
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Hi Khalid,
I think, The question talks about Compile time error.
You can run it only when the file is compiled.
 
Khalid Bou-Rabee
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Oh, I see, it automatically compiles both classes when you compile one as long as they're in the same directory...
if this happens, then you'll get a compile error with the protected method not being able to be seen out of it's package...

Argh, TRICKY!
Khalid
 
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Originally posted by Khalid Bou-Rabee:

if this happens, then you'll get a compile error with the protected method not being able to be seen out of it's package...
Argh, TRICKY!
Khalid


You still didn't get it !
The error is bcoz the file Base is searched for in the package in which the compilation unit Class1.java exists.
Since it cannot find it, it gives the above error 4)
Hope this helps !
 
Anonymous
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Originally posted by shikhar:
Hi
Pl have a look at the undergiven code.
The correct answer is 4.
Can anybody explain this error.

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 answer is 4, because the modifier of class Base is "default" ,this means only the classes in the same package Base can access it. The class Class1 is in package Class1, so it can't access class Base.
 
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