Static initializer order

[Chris Allen]

Here is a question from the Whizlabs free trial for SJCP 1.4 exam:

public class Static {
  static {
    int x=5;
  }
 
  static int x,y;
  public static void main(String [] args) {
    x--;
    myMethod();
    System.out.println(x+y+ ++x);
  }
 
  public static void myMethod() {
    y=x++ + ++x;
  }
}

I am confused as to why the answer is 3. The answer indicates that the values of x and y are set to zero. Why wouldn't x be set to 5?

[Corey McGlone]

Take special note of the line in the static initializer:

int x=5;

See the "int" at the beginning of that line? Due to that, we don't assign 5 to the class variable, we create a new local variable and assign 5 to that. The class variable, Static.x is initialized to 0.

Then, to start the main method, we decrement x, taking it to -1.

When we invoke myMethod, we get this:

y=x++ + ++x;  // y = 0, x = -1
y= -1 + ++x;  // y = 0, x = 0
y= -1 +  1;   // y = 0, x = 1

So, when the assignment is applied x is equal to 1 and y is set to 0.

Finally, back in the main method, we execute this line:

System.out.println(x+y+ ++x);  // x = 1, y = 0
System.out.println(1+0+  2 );  // x = 2, y = 0
System.out.println(3);

Hopefully, that helps you see why you get the result that you do.

[Chris Allen]

Ahhh, such a small little word but makes all the difference in the world. Thanks.