About constructor initialize sequence when using inheritance in Java

Question

I am trying to do the exercise "reusing/E07_SimpleInheritance2" in "Thingking in Java 4th edition". the code works but the output in the console is:

A: New instance C
B: New instance B
C: New instance C

But I think C should be in front of B because the sentence "System.out.println('C: ' + str);" is in the C2's constructor, followed by instance B. Well, why not?

OK, I just realized when initialize, the sequences is: (Static variables, static fields) > (variables, fields) > Constructors. That's the reason. Problem solved, thanks for the guys bellow :)

class A2{
    A2(String str){
        System.out.println("A: " + str);
    }
}

class B2{
    B2(String str){
        System.out.println("B: " + str);
    }
}

class C2 extends A2{
    C2(String str){
        super(str);
        System.out.println("C: " + str); //I think it should work first
    }
    B2 b = new B2("New instance B"); //Then followed by B
}
public class Q7_7_SimpleInheritance2 {

    public static void main(String[] args) {
        C2 c = new C2("New instance C");

    }

}

@DawoodibnKareem Ooops, I misread. Thought his problem was why "A" came before "B". — daniu
– daniu, Commented Jun 19, 2018 at 11:05

Eran · Accepted Answer · 2018-06-19 09:55:10Z

4

When a C2 instance is created, the first thing to happen is executing the A2 (super class) constructor, which results in A: New instance C being printed first.

Then the instance variables of C2 are initialized, before the body of C2's constructor. Therefore B2 b = new B2("New instance B"); is executed before System.out.println("C: " + str);, so B: New instance B is the second output line and C: New instance C the final line.

You may have been confused by the instance variable b being declared after the constructor, but that has no meaning.

Changing the code as follows may be less confusing, but would yield the exact same output:

class C2 extends A2 
{
    B2 b = new B2("New instance B");
    C2(String str){
        super(str);
        System.out.println("C: " + str);
    }
}

edited Jun 19, 2018 at 9:55

answered Jun 19, 2018 at 9:54

Eran

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2 Comments

Arnaud Over a year ago

And the relevant JLS link : docs.oracle.com/javase/specs/jls/se7/html/jls-12.html#jls-12.5

Edward Over a year ago

Yes. I just realized when initialize, the sequences is: (Static variables, static fields) > (variables, fields) > Constructors. So whatever the b being declared...

NiVeR · Accepted Answer · 2018-06-19 12:42:10Z

2

The key point here is that when instantiating a class the following happens(in this order):

super is called
class members are initialized
other constructor statements are executed

2 and 3 explain why B: New instance B is printed before C: New instance C.

edited Jun 19, 2018 at 12:42

answered Jun 19, 2018 at 9:56

NiVeR

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Comments

Kimses · Accepted Answer · 2018-06-19 09:57:30Z

-1

A is before C because the constructor of C calls the constructor of A before writing to sysout.

answered Jun 19, 2018 at 9:57

Kimses

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1 Comment

Dawood ibn Kareem Over a year ago

The question was why is B before C.

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