Inheritance
amarkirt saroay
Ranch Hand
Posts: 167
posted 9 years ago
public class Test{
public static void main(String []kk){
int y=9;
long b=9;
A obj=new B();
obj.meth(y);
new B().meth(b);
}
}
class A{
public void meth(long u){
System.out.println("parent");
}
}
class B extends A{
public void meth(int aa)
{
System.out.println("child");
}
}
Result is:
parent
parent
public static void main(String []kk){
int y=9;
long b=9;
A obj=new B();
obj.meth(y);
new B().meth(b);
}
}
class A{
public void meth(long u){
System.out.println("parent");
}
}
class B extends A{
public void meth(int aa)
{
System.out.println("child");
}
}
Result is:
parent
parent
why in both cases the parent version of meth gets called?
SCJP-75%
SCWCD-82%
posted 9 years ago
Note that you're not overriding here. You're overloading. Class B inherits meth(long) from A, and also overloads with its own meth(int).
"We're kind of on the level of crossword puzzle writers... And no one ever goes to them and gives them an award." ~Joe Strummer
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Srividhya Kiran
Ranch Hand
Posts: 166
posted 9 years ago
Hi,
When methods are overloaded reference type matters. So call to obj.meth(y); will call the super class because the reference type is of the superclass. y is a integer variable so widening occurs to long and you get the output "parent".
Now in this call to new B().meth(b); the variable you pass to the meth() is long and since subclass B has both the versions of meth it will call the class A meth() that takes long as an argument.
Hope this helps you
Srividhya
When methods are overloaded reference type matters. So call to obj.meth(y); will call the super class because the reference type is of the superclass. y is a integer variable so widening occurs to long and you get the output "parent".
Now in this call to new B().meth(b); the variable you pass to the meth() is long and since subclass B has both the versions of meth it will call the class A meth() that takes long as an argument.
Hope this helps you
Srividhya
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