Analyze the following code: public class Test { public static void main(String[] args) { int[] x = new int[5]; int i; for (i = 0; i < x.length; i++) x[i] = i; System.out.println(x[i]); } } Group of answer choicesThe program displays 0 1 2 3 4.The program displays 4.The program has a runtime error because the last statement in the main method causes ArrayIndexOutOfBoundsException.The program has a compile error because i is not defined in the last statement in the main method.
Question
Analyze the following code: public class Test { public static void main(String[] args) { int[] x = new int[5]; int i; for (i = 0; i < x.length; i++) x[i] = i; System.out.println(x[i]); } } Group of answer choicesThe program displays 0 1 2 3 4.The program displays 4.The program has a runtime error because the last statement in the main method causes ArrayIndexOutOfBoundsException.The program has a compile error because i is not defined in the last statement in the main method.
Solution
The program has a runtime error because the last statement in the main method causes ArrayIndexOutOfBoundsException.
Here's the step-by-step analysis:
- The code declares a class named Test.
- Inside the Test class, it defines the main method which is the entry point of any Java application.
- Inside the main method, it declares an array x of integers with a size of 5. So, the valid indices of the array are 0 to 4.
- Then, it declares an integer variable i.
- The for loop initializes i to 0 and runs as long as i is less than the length of the array x (which is 5). In each iteration, it assigns the value of i to the ith element of the array x.
- After the loop, it tries to print the ith element of the array x. However, at this point, the value of i is 5 (because the loop has just finished running), which is not a valid index for the array x. Hence, it throws an ArrayIndexOutOfBoundsException at runtime.
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