How to find a integer in an array using a Generics Method

Question

I am still new to Generics and I am wondering why I can't do "if(b[m] > key)" its probably something very simple and I'm too tired to notice. But As you can see I am trying to find a certain number in an array of integers. What am I doing wrong? It has to be a generic method.

public class Question1<T>
{    
/**
 * This method searches through the array of integers to check if they match the key.
 * @param b array
 * @param key certain number
 */
public static <T> void search(T[] b,T key)
{
   int l = 0; 
    int r = b.length - 1;
    while (l <= r)
    {
        int m = (l + (r-l)/2);
        if(b[m].equals(key))
            System.out.println("Key: " + key.toString()+ " is in Element: " + m);

        if (b[m] > key)
        {
            l = m + 1;
        }
        else
        {
            r = m - 1;
        }
    }
    System.out.println("Not in array.");        
}
public static void main(String[] args)
{

    Integer[] iray = {1,2,3,4,5};
    int key = 4;
    search(iray,key);
}

Answer 1

The error occurs because T can literally be any reference type. Let's say T is now java.util.UUID or java.util.Scanner or java.util.List. Objects of these types cannot be compared by >. Some can't even be compared at all.

So what you need to do is to constrain the generic parameter T to types that can be compared:

public static <T extends Comparable<T>> void search(T[] b,T key)
                 ^^^^^^^^^^^^^^^^^^^^^

Now, you can call compareTo on b[m] to compare them. The method will return a positive result if b[m] is greater than the parameter:

if (b[m].compareTo(key) > 0)

Alternatively, you can add a new Comparator parameter to let callers of this method specify how they want the objects to be compared:

public static <T> void search(T[] b,T key, Comparator<? super T> comparator)

Then the comparison line will become:

if (comparator.compare(b[m], key) > 0)

Also, you seem to have (accidentally?) made Question generic. That is unnecessary unless you have some code not shown that is using it.