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Select the correct answerWhat will be the output of the following Java program? class access { public int x; static int y; void cal(int a, int b) { x += a ; y += b; } } class static_specifier { public static void main(String args[]) { access obj1 = new access(); access obj2 = new access(); obj1.x = 0; obj1.y = 0; obj1.cal(1, 2); obj2.x = 0; obj2.cal(2, 3); System.out.println(obj1.x + " " + obj2.y); } }Options3 22 31 21 5

Question

Select the correct answerWhat will be the output of the following Java program? class access { public int x; static int y; void cal(int a, int b) { x += a ; y += b; } } class static_specifier { public static void main(String args[]) { access obj1 = new access(); access obj2 = new access(); obj1.x = 0; obj1.y = 0; obj1.cal(1, 2); obj2.x = 0; obj2.cal(2, 3); System.out.println(obj1.x + " " + obj2.y); } }Options3 22 31 21 5

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Solution

The correct answer is 1 5.

Here's the step by step explanation:

  1. Two objects of class 'access', obj1 and obj2 are created.
  2. For obj1, x is set to 0 and y is also set to 0. Then the method cal is called with parameters 1 and 2. This adds 1 to x and 2 to y. So, obj1.x becomes 1 and y becomes 2.
  3. For obj2, x is set to 0. Then the method cal is called with parameters 2 and 3. This adds 2 to x and 3 to y. So, obj2.x becomes 2. But y becomes 5 because y is a static variable, which means it is common to all objects of the class. The previous value of y (which was 2) is added with 3, making it 5.
  4. Finally, obj1.x and obj2.y are printed, which are 1 and 5 respectively.

This problem has been solved

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