问题
When I try the code on the bottom of this question, the output is:
a a b a b c
So this means that the constructor from B and C call the constructors from their superclasses. But why? I thought that the superclass' constructor only get's called when used with the super() function like this:
public sportscar(String name, int weight, int topspeed){
super(name, weight);
this.setTopspeed(topspeed);
}
But if it automatically takes over the constructor from the classes they extend from, why would we use the super() function? I know that normal methods extend to there sub-classes automaticaly, but I thought that the constructor was different.
If anyone can clear this out for me, thanks a lot!
Code:
public class App {
public static void main(String[] args) {
new A();
new B();
new C();
}
}
public class A {
public A(){
System.out.println("a ");
}
}
public class B extends A {
public B(){
System.out.println("b ");
}
}
public class C extends B {
public C(){
System.out.println("c ");
}
}
回答1:
This behaviour is mandated by the JLS (§8.8.7. Constructor Body):
If a constructor body does not begin with an explicit constructor invocation and the constructor being declared is not part of the primordial class
Object, then the constructor body implicitly begins with a superclass constructor invocation "super();", an invocation of the constructor of its direct superclass that takes no arguments.
来源：https://stackoverflow.com/questions/23899863/java-why-does-my-class-automatically-inherits-constructor-from-superclass