java构造函数顺序 [英] java constructor order

问题描述

这个java程序很容易和充满评论，所以你可以理解它fast.however，为什么在构造staff [1]，程序首先去语句：

  this（Employee＃+ nextId，s）; 
  

然后转到对象初始化块，然后回到语句，如何confusion.why不它首先使用对象初始化块

  import java.util。*; 
 
 public class ConstructorTest 
 {
 public static void main（String [] args）
 {
 //用三个Employee对象填充staff数组
员工[] staff =新员工[3]; 
 
 staff [0] = new Employee（Harry，40000）; 
 staff [1] =新员工（60000）; 
 staff [2] = new Employee（）; 
 
 //打印出所有Employee对象的信息
 for（Employee e：staff）
 System.out.println（name =+ e.getName（）
 +，id =+ e.getId（）
 +，salary =+ e.getSalary（））; 
} 
} 
 
 class Employee 
 {
 //三个重载的构造函数
 public Employee（String n，double s）
 {
 name = n; 
 salary = s; 
} 
 
 public Employee（double）
 {
 //调用Employee（String，double）构造函数
 this（Employee＃ nextId，s）; 
} 
 
 //默认构造函数
 public Employee（）
 {
 //名称初始化为 - 见下面
 // salary not displayed set  - 初始化为0 
 //在初始化块中初始化id 
} 
 
 public String getName（）
 {
 return名称; 
} 
 
 public double getSalary（）
 {
 return salary; 
} 
 
 public int getId（）
 {
 return id; 
} 
 
 private static int nextId; 
 
 private int id; 
 private String name =; //实例字段初始化
 private double salary; 
 
 //静态初始化块
 static 
 {
 Random generator = new Random（）; 
 //将nextId设置为0到9999之间的随机数
 nextId = generator.nextInt（10000）; 
} 
 
 //对象初始化块
 {
 id = nextId; 
 nextId ++; 
} 
} 
  

解决方案

this（Employee＃+ nextId，s）; 包括对超类构造函数的隐式调用，当然必须在子类的初始化程序块之前执行。 p>

使用实例初始化器通常是一个坏主意，因为它们不是众所周知的，不能做构造函数，混合两者会导致混乱。

This java program is easy and full of comment,so you can understand it fast.however,why in construct staff[1],the program first go to the statement:

this("Employee #" + nextId, s);

then go to the object initialization block,and then go back to the statement,how confusion.why not it first use the object initialization block

import java.util.*;

public class ConstructorTest
{
   public static void main(String[] args)
   {
      // fill the staff array with three Employee objects
      Employee[] staff = new Employee[3];

      staff[0] = new Employee("Harry", 40000);
      staff[1] = new Employee(60000);
      staff[2] = new Employee();

      // print out information about all Employee objects
      for (Employee e : staff)
         System.out.println("name=" + e.getName()
            + ",id=" + e.getId()
            + ",salary=" + e.getSalary());
   }
}

class Employee
{
   // three overloaded constructors
   public Employee(String n, double s)
   {
      name = n;
      salary = s;
   }

   public Employee(double s)
   {
      // calls the Employee(String, double) constructor
      this("Employee #" + nextId, s);
   }

   // the default constructor
   public Employee()
   {
      // name initialized to ""--see below
      // salary not explicitly set--initialized to 0
      // id initialized in initialization block
   }

   public String getName()
   {
      return name;
   }

   public double getSalary()
   {
      return salary;
   }

   public int getId()
   {
      return id;
   }

   private static int nextId;

   private int id;
   private String name = ""; // instance field initialization
   private double salary;

   // static initialization block
   static
   {
      Random generator = new Random();
      // set nextId to a random number between 0 and 9999
      nextId = generator.nextInt(10000);
   }

   // object initialization block
   {
      id = nextId;
      nextId++;
   }
}

解决方案

Because this("Employee #" + nextId, s); includes an implicit call to the superclass constructor, which of course must be executed before the initializer block of the subclass.

Using instance initializers is generally a bad idea as they are not well known, cannot do anything more than constructors, and mixing both leads to confusion.

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