Why was the static initialization block executed after the constructor?

Question

Good day colleagues!

In the literature they write that when a class is loaded, the static block is first executed, and then the dynamic block and the constructor. I managed to write a class so that the static block was executed last.

public class Singleton { static Singleton st = new Singleton(); static { //статический блок System.out.println("Статический блок"); } { //анонимный блок System.out.println("Динамический блок"); } Singleton(){ //конструктор System.out.println("Конструктор"); } public static void main(String[] args) { //Пустая точка входа } }

OUTPUT:

 Динамический блок Конструктор Статический блок

Question: How to explain the observed behavior?

Accepted Answer · 2018-08-15T10:02:59

Let's take a look:

Your first static field is declared:

 static Singleton st = new Singleton();

It needs to be initialized, respectively, the constructor is called:

 Singleton() { System.out.println("Конструктор"); }

But before the constructor is called, there is a dynamic initialization block, so the object is first initialized in this block, and then the constructor is executed:

 { //анонимный блок System.out.println("Динамический блок"); } //затем вызывается непосредственно конструктор

So, the first line of the class is ready, the static field static Singleton st initialized, then we already initialize the static block. Can you, on the basis of my answer, suppose what happens if you enter another call to the Singleton constructor in a static block? Assume, and then you can try and compare your guess.

UPD.

This is not to say that the static block is called before the dynamic block. They are called under different conditions. The static block is called after the class loader is loaded by the class, and the dynamic block is called when the class is instantiated . That is static - class initialization; dynamic - class instance initialization. And the dynamic block in your particular example is called during the initialization of the Singleton instance in the first line, which is a static field.

if static -> dynamic -> constructor were called, everything would be clear to me.
Maybe explain why static is not called before the dynamic block?
After all, if you remove the static field, and create an object in the Maine, the order will be exactly this: static -> dynamic -> constructor.
Initially, static variables are initialized, then static blocks, etc. ...
@OleksiyMorenets So why the static block was not initialized in the first place?
What, when initializing a statistical field, do you need not to perform a static block at all?
Change the static block and static variable in some places and you will be very surprised :)
Rather, to say that the program began with the initialization of a static field, which contains an object that needs to be initialized.

Johann stolz johann stolz 66 four · Answer 2 · 2018-08-15T11:11:57

Initialization of static fields and static blocks are performed in the order they are declared in the class. Since you have static static field static Singleton st = new Singleton(); is declared first, it is initialized first. When a field is initialized, a class object is created, but the execution of static block when creating an object is skipped, because its execution will start only after initializing the static field declared first. Accordingly, when initializing this field, the static block is ignored, the Dynamic block is executed, then the class constructor, the field is now initialized, and, in accordance with the declaration order, the static block is executed. http://www.quizful.net/post/java-fields-initialization

Why was the static initialization block executed after the constructor?

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