Java Constructors - Order of execution in an inheritance hierarchy

Question

Consider the code below

  class Meal {
    Meal() { System.out.println(\"Meal()\"); }
  }

  class Bread {
    Bread() { System.out.println(\"Bread()\"); }
          


        

Answer 1

These are instance fields

private Bread b = new Bread();
private Cheese c = new Cheese();
private Lettuce l = new Lettuce();

They only exist (execute) if an instance is created.

The first thing that runs in your program is

public static void main(String[] args) {
     new Sandwich();
}

Super constructors are called implicitly as the first thing in each constructor, ie. before System.out.println

class Meal {
    Meal() { System.out.println("Meal()"); }
}

class Lunch extends Meal {
    Lunch() { System.out.println("Lunch()"); }
}

class PortableLunch extends Lunch {
    PortableLunch() { System.out.println("PortableLunch()");}
}

After the super() calls, instance fields are instantiated again before the constructor code.

The order, reversed

new Sandwich(); // prints last
// the instance fields
super(); // new PortableLunch() prints third
super(); // new Lunch() prints second
super(); // new Meal(); prints first