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Practice exam question

 
Greenhorn
Posts: 8
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This question was on the Jxam simulator:
What is displayed when the following piece of code is compiled and executed:
class Test{
public static void main(String [] args){
Base b = new Subclass();
System.out.println(b.x);
System.out.println(b.method());
}
}
class Base{
int x = 2;
int method(){
return x;
}
}
class Subclass extends Base{
int x = 3;
int method(){
return x;
}
}
The output when executed is :
2
3
It explained that with the line: Base b = new Subclass();
calling varibles of b will come from the base class and methods will come from the Subclass?
This is hard to get a handle on. Any insite, or just memorize instead of rationalize?? Thanks


------------------
 
Author and all-around good cowpoke
Posts: 13078
6
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The key point here is that variable references are resolved at compile time, while method references are resolved at run time.

Base b = new Subclass();
System.out.println(b.x); // resolved to the Base class address
System.out.println(b.method()); // resolved to the actual object method
Bill
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Ranch Hand
Posts: 2379
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Hi Iarry,
see some previous post on this topic with the subjects like "Dunamic method lookup" or "Method overriding" or "Variable shadowing" in this forum. We have already discussed on ur query a lot. These r tricky topics indeed....

------------------
azaman
 
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