任务02：此Java代码如何工作？

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.
    }
}

Okay guys, as seen above I have got a new Java Code, I think all in all I did understand many things, at least I did comments above the stuff I understood and thought my way of thinking would be correct.

Feel free and check my above comments and correct me if I am wrong.

I have tested and the above code actually prints following:

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

So in other words I was partially right, I simply don't understand why it is printing test: 07, since there is no loop, and why t2: 87 is not t2: 86?

I was initially expecting something like that:

test: 03
t1: 03
t2: 86

Any profis wanting to check it?

Thanks in advance.

Answer 1

Learning is called every time you create a new Learning() , because it's the constructor for all instances of the Learning class. You create two instances of Learning , therefore test gets printed twice. The reason the value is 87 is because s is static . This means all instances of Learning share the same value for s . Therefore modifying t1 's instance of s to be 6 also modifies t2's instance of s , which then gets incremented in its constructor, and becomes 7 .

Answer 2

It's printing test: 07 because you are instantiating a new copy of Learning() as t2. Therefore, it's going in the constructor and printing the line in there.

Because it's being instantiated, this also increases the value of s in the constructor. Because s is a static variable, its value is shared among all Learning objects. Because you had set it to 6 in t1, it increases to 7 when you instantiate the new object, yielding '87'.

Answer 3

Since you create 2 objects of the class learning, the "test: 0x" will print twice.

The reason why t2 is 86 and not 87 is because you first set t1.s = 6 and, after that, call the constrctor new Learning() , which will increment s by 1.

Answer 4

Look at this line: Learning t2 = new Learning();

This will cause test: 07 to be printed out because it uses the Learning() constructor which has a print statement in it.

This line also causes s to increase by 1 from 6 to 7. This is why t2:87 is printed instead of t2:86

Answer 5

Every time you create a new instance of Learning type, the constructor will be invoked. So when you create the t2 object the s value is incremented by one(that's why you have the 87 value printed instead of 86 and the text test: 07 is printed.