java - 任务02 : How does this Java Code work?

Question

public class Learning {
    int i = 1; // i is assigned to 1.
    static int s = 2;// s is assigned to 2.

    Learning() {
        i = 0; s++; // i is assigned to 0 and s increases to 2. 
        System.out.println("test: " + i + s);// this line will display test: 03 since the values are above given.
    }

    public static void main (String [] args) {
        Learning t1 = new Learning();// Creating and instance variable from the Class Learning.
        System.out.println("t1: " + t1.i + t1.s);//t1 reaches with the help of the . the value of i as well as s and again this line will print out t1: 03
        t1.s = 6;// Now the variable get s new value which is 6
        Learning t2 = new Learning();// Creating another instance variable from the class called Learning
        t2.i = 8;// Now the value for i is set for 8.
        System.out.println("t2: " + t2.i + t2.s);// Now I thought the program would display t2: 86 since I have already got the value 8 and 6 assigned above, but it doesn't do that.
    }
}

好吧伙计们，正如上面所看到的，我有了一个新的Java代码，我想总而言之我确实理解了很多东西，至少我在我理解的东西上面做了注释，并且认为我的思维方式是正确的。

请随意检查我的上述评论，如果我错了，请纠正我。

我已经测试过，上面的代码实际打印如下:

test: 03
t1: 03
test: 07
t2: 87

所以换句话说我是部分正确的，我根本不明白为什么它打印 test: 07，因为没有循环，为什么 t2: 87 不是 t2: 86？

我最初期待这样的事情:

test: 03
t1: 03
t2: 86

有专业人士想检查一下吗？

提前致谢。

Answer 1

每次创建 new Learning() 时都会调用

Learning，因为它是 Learning 类的所有实例的构造函数。您创建了两个 Learning 实例，因此 test 被打印两次。该值为 87 的原因是 s 是 static。这意味着 Learning 的所有实例都共享相同的 s 值。因此，将 t1 的 s 实例修改为 6 也会修改 t2 的 s 实例，然后该实例会递增在其构造函数中，变为 7。