I found a class like this:
public class Computer implements Serializable {
private static final long serialVersionUID = 1L;
//...
public Computer() {
super();
}
//...
}
Can someone explain me, how this works? The class isn't inheriting any super class and there could still be "super();" in the constructor?
By default all classes inherit java.lang.Object
. So a hidden code in your class is
public class Computer extends java.lang.Object implements Serializable {
private static final long serialVersionUID = 1L;
//...
public Computer() {
super(); //here Object's default constructor is called
}
//...
}
If a parent class has default constructor (no argument) and if a child class defines a default constructor, then an explicit call to parent's constructor is not necessary.
Java does it implicitly, in other words, Java puts super()
before first statement of the child's constructor
consider this example
class Base {
public Base() {
System.out.println("base");
}
}
class Derived extends Base {
public Derived() {
//super(); call is not necessary.. Java code puts it here by default
System.out.println("derived");
}
}
So when you create Derived d = new Derived();
the output is..
base
derived
When you do not extends
any class then by default the inherited class is Object
so it calls Object
class constructor.
In Java, Object
class is at the top of the Class
hierarchy. Implicitly, All the Classes are implicitly of Object
type. Invocation of super()
method is not needed explicitly in code. But, this will be added by the compiler during conversion of Class into bytecode. so, the first line in any constructor would be invocation of super()
implicitly by default. Construction of an object includes all the invocation of default constructors of Super classes. But, you can also call a customized argument constructor of the super
class in your class
constructor.
By default every java class implicitly inherits java.lang.Object
class. So, super()
will call the Object
class constructor in your case!
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