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class TestArray {    public static void main(String args[]) {      int arr_sample[] = new int[2];      System.out.println(arr_sample[0]);    }}What will be the result of compiling and executing the above code?Select one:a.The program does not compile because arr_sample[0] is being read before being initialized.b.The program compiles and prints 1 when executed.c.The program compiles and prints 0 when executed.d.The program generates a runtime exception because arr_sample[0] is being read before being initialized.e.The program compiles and runs but the results are not predictable because of un-initialized memory being read.

Question

class TestArray {    public static void main(String args[]) {      int arr_sample[] = new int[2];      System.out.println(arr_sample[0]);    }}What will be the result of compiling and executing the above code?Select one:a.The program does not compile because arr_sample[0] is being read before being initialized.b.The program compiles and prints 1 when executed.c.The program compiles and prints 0 when executed.d.The program generates a runtime exception because arr_sample[0] is being read before being initialized.e.The program compiles and runs but the results are not predictable because of un-initialized memory being read.

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Solution

The correct answer is c. The program compiles and prints 0 when executed.

Here's why:

In Java, when you declare an array, each element in the array is automatically initialized with a default value. The default value depends on the type of the array. For an array of integers, like in this case, the default value is 0.

So, when the code tries to print the value of arr_sample[0], it prints 0, because that's the default value that was automatically assigned to this element of the array when the array was declared.

This problem has been solved

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