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Siva Nookala - 11 Apr 2016
 Since array contains multiple elements, but we have only one variable, we use index to access the elements. The index is nothing but an unique number which identifies each element. Using the index, the element at that index can be initialized or can be read. The index has to be specified within square brackets and all array indexes start at zero. The index of first element is zero, second element is 1 etc., So, for accessing fifth element in an array `student_marks`, you need to use `student_marks = 87;` The fifth element is accessed using index four (4). Similarly tenth element is accessed using index 9 and 8th element is accessed using 7. For an array of size N, the valid indices are `0, 1, 2 ..., N-2, N-1`. For array with size 4, the valid indices are 0, 1, 2, 3. If any other index is used to access the element then `ArrayIndexOutOfBoundsException` will be thrown. Student Marks ArrayCODE Try it Online`class StudentMarksArray{    public static void main(String arg[])    {        int student_marks[] = new int;         student_marks = 25; // LINE A        System.out.println("Third Element = " + student_marks); // LINE B        System.out.println("Fourth Element = " + student_marks); // LINE C                // student_marks[-3] = 45; // Won't work // LINE D        // student_marks = 32; // Won't work // LINE E        }}`OUTPUTThird Element = 25Fourth Element = 0DESCRIPTIONIn LINE A, we are assigning 25 to the third element i.e. index 2. In LINE B, we are printing third element i.e. also index 2. So it will print the previously assigned value, which is 25. In LINE C, we are printing fourth element i.e. index 3. Since it was not assigned previously, the original default value of 0 is printed.THINGS TO TRYUncomment `LINE D` - to see the `ArrayIndexOutOfBoundsException` Similarly uncomment `LINE E` - to see the `ArrayIndexOutOfBoundsException`.
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