Java parameter passing with the values passed

Reference: https://blog.csdn.net/zhenge1990/article/details/78897166

Java method of delivery, there are two parameters, and the value is passed reference.

1. passed by value:

Parameter type int, long, etc. eight basic data types, the process parameters are passed by value of the copy process

The following code

 public  static  void main (String [] args) {
         int A =. 5 ; 
        Fun (A); 
        System.out.println (A); // output result. 5 
    } 
 
    Private  static  void Fun ( int A) { 
        A +. 1 = ; 
    }

2. passed by reference

Type process parameter reference type, parameters passed by way of reference copy

public class Test {
 
    public static void main(String[] args) {
        A a = new A(5);
        fun(a);
        System.out.println(a.a);// 输出结果为6
    }
 
    private static void fun(A a) {
        a.a += 1;
    }
 
    static class A {
        public int a;
 
        public A(int a) {
            this.a = a;
        }
    }
}

　　Look at the following situation:

public  class the Test { 
 
    public  static  void main (String [] args) { 
        Integer A =. 5 ; 
        Fun (A); 
        System.out.println (A); // output result. 5 
    } 
 
    Private  static  void Fun (Integer A) { 
        A + =. 1 ; 
    } 
 
}

Here obviously is passed by reference, why not change the value of the object of it?

Here the basic data types actually used autoboxing function package type.

Integer a = 5, actually compiled Integer a = Integer.valueOf (5), see Integer source, does not change the value of the original object, but which is a reference to point to another object.