What is the code will print out? And why?

Question

public class test {
int x = 0;
test() {
x = 3;
System.out.println(x);
}
test (int y) {
x = y+2;
System.out.println(x);
}
void doSomething(int y) {
x += y;
System.out.println(x);
}
void doMore(int y) {
System.out.println(x);
doSomething(y+1);
}
}

public class mainProgram {
public static void main(String[] args) {
test obj1 = new test();
obj1.doSomething(1111);
obj1.doMore(1111);
test obj2 = new test(1111);
}
}

Answer 1

test default constructor gets called for initialization of object. 3 is assigned to and x is printed.

3

doSomething is called with argument 1111. 1111 is added to x and assigned to x. x is printed.

1114

doMore is called with argument 1111, prints x and calls doSomething with argument 1112.

1114

doSomething is called with argument 1112, and 1112 is added to x and assigned to x. x is printed.

2226

test constructor with arguments is called during initialization of obj2 with argument 1111. 2 is added to 1111 and assigned to x. x is printed.

1113

Answer 2

... If you can remove the spaces in between lines, that would be great.

Answer 3

Joseph Jostar When you do

test obj1 = new test();

You are calling the default constructor (defined on line 3) which only assigns 3 to x. When you do

test obj2 = new test(1111);

the program will call the constructor defined on line 7 to add 1111 to 3 and assign the result to x.

You are correct for the part about doSomething.

Answer 4

Hatsy Rei thank you
i still have a few questions.
1. why test obj1 = new test();
did not print out 1113?
2. when doSomething(y+1) called
it’s x= 1114? So, it’s like x=1114+1111+1

Thank you

Answer 5

Hatsy Rei thank you for your help and wish you have a great weekend ：）

Answer 6

Hatsy Rei i will