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Consider a Program:public class Test {            public static void main(String[] args) {             int[] numbers = {1, 2, 3, 4, 5};             for (int i = 0; i <= numbers.length; i++) {             System.out.println(numbers[i]);             }            }}Which of the following is an output of the above program ? a. It will print numbers from 1 to 5 b. It will print numbers from 1 to 5 and then throw an exception c. It will compile but not run d. It will not compile

Question

Consider a Program:public class Test {            public static void main(String[] args) {             int[] numbers = {1, 2, 3, 4, 5};             for (int i = 0; i <= numbers.length; i++) {             System.out.println(numbers[i]);             }            }}Which of the following is an output of the above program ? a. It will print numbers from 1 to 5 b. It will print numbers from 1 to 5 and then throw an exception c. It will compile but not run d. It will not compile

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Solution

The output of the above program will be option b. It will print numbers from 1 to 5 and then throw an exception.

Here's the step by step explanation:

  1. The program declares an array of integers named 'numbers' with elements 1, 2, 3, 4, 5.

  2. It then enters a for loop where the variable 'i' is initialized to 0.

  3. The loop condition is 'i' less than or equal to the length of the array 'numbers'. The length of the array 'numbers' is 5, so the loop will iterate from 'i' equals 0 to 5.

  4. Inside the loop, it prints the 'i'th element of the array 'numbers'.

  5. For 'i' equals 0 to 4, it will print the numbers 1 to 5 respectively.

  6. However, when 'i' equals 5, it will try to access the 6th element of the array 'numbers' which does not exist. In Java, array indices start from 0 and go up to length-1. So, trying to access an element beyond length-1 will result in an ArrayIndexOutOfBoundsException.

So, the program will print numbers from 1 to 5 and then throw an exception.

This problem has been solved

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