java - Questions about implementing the Comparable interface -

i teaching java programming class first time. textbook uses comparable, see examples on net use comparable<t>. these using 2 different interfaces?

related this, when write compareto() method in student class, textbook uses

public int compareto(object other){     if (! (other instance of student))         throw new illegalargumentexception("parameter must student");     return name.compareto( ((student)other).getname() ); } 

i see other examples of compareto() this:

public int compareto(student otherstudent){     return name.compareto( otherstudent.getname() ); } 

is second construct 1 should use if student class implements comparable<student>?

finally, textbook gives hints on how write general-purpose object sorting algorithm. arrived @ following. works, gives "unchecked or unsafe operations" warning. there way write "safe" general-purpose sorting algorithm?

private static void bubblesortobjects(comparable[] array) {     int = 0;     boolean swapped = true;     while (swapped && < array.length - 1)     {         i++;         swapped = false;         (int j = 0; j < array.length - i; j++)         {             if (array[j].compareto(array[j + 1]) > 0)             {                 comparable temp = array[j];                 array[j] = array[j + 1];                 array[j + 1] = temp;                 swapped = true;             }         }     } } 

thanks or advice.

just help, modern general purpose sorting algorithm be:

private static <t extends comparable<? super t>> void bubblesortobjects(t[] array) {     int = 0;     boolean swapped = true;     while (swapped && < array.length - 1)     {         i++;         swapped = false;         (int j = 0; j < array.length - i; j++)         {             if (array[j].compareto(array[j + 1]) > 0)             {                 t temp = array[j];                 array[j] = array[j + 1];                 array[j + 1] = temp;                 swapped = true;             }         }     } }