What is the result of the following program?1: public class Squares {2: public static long square(int x) {3: long y = x * (long) x;4: x = -1;5: return y;6: }7: public static void main(String[] args) {8: int value = 9;9: long result = square(value);10: System.out.println(value);11: } }*A. -1B. 9C. 81D. Compiler error on line 9.E. Compiler error on a different line.
Question
What is the result of the following program?1: public class Squares {2: public static long square(int x) {3: long y = x * (long) x;4: x = -1;5: return y;6: }7: public static void main(String[] args) {8: int value = 9;9: long result = square(value);10: System.out.println(value);11: } }*A. -1B. 9C. 81D. Compiler error on line 9.E. Compiler error on a different line.
Solution
The result of the program will be B. 9.
Here's the step by step explanation:
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The main method is called, and an integer variable 'value' is declared and initialized to 9.
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The 'square' method is called with 'value' as the argument.
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Inside the 'square' method, a long variable 'y' is declared and initialized to the square of 'x' (which is 'value' passed from the main method, i.e., 9). So, 'y' becomes 81.
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The variable 'x' is then set to -1. However, this does not affect the 'value' in the main method because in Java, primitive types are passed by value, not by reference. This means that changes to 'x' inside the 'square' method do not affect 'value' in the main method.
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The 'square' method then returns 'y', which is 81. But this returned value is not printed or stored anywhere in the main method.
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Back in the main method, 'value' is printed, which is still 9.
So, the output of the program is 9.
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