Code block static code block initializes the execution order of the constructor in-place

Problem Description:

Java instantiates a class. If this class has not only a constructor, but also a code block, a static code block, and an instance of another class that is initialized in place, what is the order of execution of the code?

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Samples and running results

people class

package Extends;

import Dog.Dog;

class people {
    
    
    public int age=100;
    public String name;

    Dog dog=new Dog();

    public people() {
    
    
        System.out.println("people");
    }
    {
    
    
        System.out.println("people的代码块");
    }
    static {
    
    
        System.out.println("people中的静态代码块");
    }
}

dog class

public class Dog {
    
    
    String name;
    int age;
    public Dog(){
    
    
        name="大黄";
        age=3;
        System.out.println("dog的初始化");
    }
}

Main function

public class Main {
    
    
    public static void main(String[] args) {
    
    
        people p1=new people();
        System.out.println("---------");
        people p2=new people();
    }
}

Cause Analysis:

Here, in the people class, there are not only code blocks, static code blocks, and construction methods, but also initialize a dog instance on the spot. From the results, we can see that the execution of the static code block is always the earliest, but once executed, the next level is : In-place initialization and code blocks. Whoever writes the two will execute the first, and the last one is the construction method.

import Dog.Dog;

class people {
    
    
    public int age=100;
    public String name;
    public people() {
    
    
        System.out.println("people");
    }
    {
    
    
        System.out.println("people的代码块");
    }
    static {
    
    
        System.out.println("people中的静态代码块");
    }
    Dog dog=new Dog();
}

At this time, the construction of dog is after the code block.