object question
The following code from a MCQ
class hasstatic{
privatestaticint x=100;
publicstaticvoid main(String args[]){
hasstatic hs1=new hasstatic();
hs1.x++;//1
hasstatic hs2=new hasstatic();
hs2.x++;//2
hs1=new hasstatic();
hs1.x++;//3
hasstatic.x++;//4
System.out.println("x="+x);//5
}
}
When i compile and run the code i got the answer as x=104.
"But when i dry run the code i thought the result will be 101 because line 4 code will increase the original x value which will be printed in line 5 and line 1,2,3 will not effect the x because they are seperate objects which will have seperate variable X."
Can anyone give me a explanation for what realy happening inside the code and why my assumption is wrong.
[1435 byte] By [
ram102125a] at [2007-11-27 5:59:51]
> because they are seperate objects which will have seperate variable X.
They are all separate objects, but they share the int x. That's what the "static" means. So hs1.x++, hs2.x++ and hasstatic.x++ are all incrementing the same variable.
(Note: Capitalisation is important. HasStatic. Also expressions like hs1.x++ are not very nice when HasStatic.x++ is available. If the MCQ did not make these points clear, it is most cr@ppy indeed.)
because x is a static. static means we can access those method or variable with out reference. so variable x share same memory lacation.
then when you run that code:
class hasstatic{
private static int x=100;
public static void main(String args[]){
hasstatic hs1=new hasstatic();
hs1.x++; //here x=101
hasstatic hs2=new hasstatic();
hs2.x++; //here x=102
hs1=new hasstatic();
hs1.x++;//here x=103
hasstatic.x++;//here x=104
System.out.println(x) //print 104
System.out.println(hs1.x) //also print 104
System.out.println(hs2.x)// also print 104
}
}
but if you write like this it will print 101
class hasstatic{
private int x=100; //remove static;
public static void main(String args[]){
hasstatic hs1=new hasstatic();
hs1.x++;//1
hasstatic hs2=new hasstatic();
hs2.x++;//2
hs1=new hasstatic();
hs1.x++;//3
hasstatic.x++; //Error
System.out.println(x) ;//Compile Error because you can't access normal variable into static method with out reference
System.out.println(hs1.x); //print 101
System.out.println(hs2.x);//print 101
}
}
