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Select the correct answerWhat will be the output of the following Java program? class access { static int x; void increment() { x++; } } class static_use { public static void main(String args[]) { access obj1 = new access(); access obj2 = new access(); obj1.x = 0; obj1.increment(); obj2.increment(); System.out.println(obj1.x + " " + obj2.x); } }Options 2 21 11 2Compilation Error

Question

Select the correct answerWhat will be the output of the following Java program? class access { static int x; void increment() { x++; } } class static_use { public static void main(String args[]) { access obj1 = new access(); access obj2 = new access(); obj1.x = 0; obj1.increment(); obj2.increment(); System.out.println(obj1.x + " " + obj2.x); } }Options 2 21 11 2Compilation Error

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Solution

The correct answer is 2 2.

Here's the step by step explanation:

  1. Two objects of the class 'access', obj1 and obj2, are created.
  2. The variable 'x' is static, which means it is shared among all objects of the class.
  3. obj1.x is set to 0.
  4. The increment() method is called on obj1, which increases 'x' by 1. So, 'x' becomes 1.
  5. The increment() method is called on obj2, which again increases 'x' by 1. So, 'x' becomes 2.
  6. When we print obj1.x and obj2.x, we get 2 2 because 'x' is shared among all objects of the class 'access'.

This problem has been solved

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