Exam 1z0-808 topic 1 question 202 discussion

super can be called at any time in a method, but inside a constructor it must be called called first. Good answer is A

Correct is A. Tested

The correct answer is A. it will print Child10 Child100 Parent100

A is correct, tested

The answer is A, I tested. But why don't we get compilation error as super and this are supposed to be first statement in a constructor ? Can someone explain?

Answer is A , straight forward

Answer is A. UAK94 provided us the source code, in their comment.

Answer is A. class S1 { protected void display(int x) { System.out.println("Parent" + x); } } public class S2 extends S1 { public void display(int x, int y) { this.display(x); display(y); super.display(y); } public void display(int x) { System.out.println("Child" + x); } public static void main(String[] args) { S2 sobj = new S2(); sobj.display(10, 100); // Child10 Child100 Parent100 } }

Answer should be A. The Main method not found is just the omitted part of this question.

The correct answer is letter D, a super has to be instantiated first in a method.