Default constructor

[Meena R. Krishnan]

source : Enthuware

public class SubClass extends SuperClass
{
     int i, j, k;
     public SubClass( int m, int n )     {  i = m ;  j = m ;  } //1
     public SubClass( int m )  {   super(m );   } //2
 }

Which of the following constructors MUST exist in SuperClass for SubClass to compile correctly?


given answers:
public SuperClass(int a);
public SuperClass();


Why do we need the default constructor in this case? The only constructor that the subclass uses of superclass is the one that takes an int. Not sure why we need the no-arg constructor for the subclass point of view.

[Meena R. Krishnan]

Got it!
The subclass constructor SubClass(int, int), since it is not explicitly calling the super class constructor, a default no-arg constructor of the superClass will be inserted in order for it to get constructed and that's why the need for the no-arg constructor.

[krishna bulusu]

Thats great man!!!

[marc weber]

Originally posted by M Krishnan:
... The subclass constructor SubClass(int, int), since it is not explicitly calling the super class constructor, a default no-arg constructor of the superClass will be inserted in order for it to get constructed and that's why the need for the no-arg constructor.

Just to be clear: Without an explicit call to super (or this), an implicit call to the superClass no-args constructor will be inserted in the subclass constructor.

The compiler will add a default no-args constructor to the superClass only if no other constructors are defined. Otherwise, it is left to the programmer to write the no-args constructor.