What will be printed after executing following program code?
class Base{
int value = 0;
Base(){
addValue();
}
void addValue(){
value += 10;
}
int getValue(){
return value;
}
}
class Derived extends Base{
Derived(){
addValue();
}
void addValue(){
value += 20;
}
}
public class Test{
public static void main(String[] args){
Base b = new Derived();
System.out.println(b.getValue());
}
}
class Base{
int value = 0;
Base(){
addValue();
}
void addValue(){
value += 10;
}
int getValue(){
return value;
}
}
class Derived extends Base{
Derived(){
addValue();
}
void addValue(){
value += 20;
}
}
public class Test{
public static void main(String[] args){
Base b = new Derived();
System.out.println(b.getValue());
}
}A. 30
B. 10
C. 40
D. 20
E. None of these
Answer: Option C
Solution (By Examveda Team)
When object of new derived is called, the flow goes to Derived() first, by default super(); is present in Derived() as the first statement, so the flow now goes to Base. Here value is initialised to 0 and then addValue() is called. The addValue has been overridden in Derived() hence The Base's addValue() will perform value+20(0+20) .After this control flows back to Derived()'s addValue() where again value+20 is done (20+20). Hence Answer is 40This Question Belongs to Java Program >> Inheritence
10+20 how 40
Sir I don't know how the output 40 please tell me the answer
When object of new derived is called, the flow goes to Derived() first, by default super(); is present in Derived() as the first statement, so the flow now goes to Base. Here value is initialised to 0 and then addValue() is called. The addValue has been overridden in Derived() hence The Base's addValue() will perform value+20(0+20) .After this control flows back to Derived()'s addValue() where again value+20 is done (20+20). Hence Answer is 40
i dont know how the output is 40.. please anyone explain that