Comparable

[Vijay Chandran]

    
public class Test 
{
    public <T extends Comparable> T findLarger(T x , T y)
    {
        if(x.compareTo(y) > 0)
        {
            return x;
        }
        else
        {
            return y;
        }
    }
    public static void main(String[] args)
    {
        Test t = new Test();
        int x1 = t.findLarger(123, new Integer(456));
        System.out.println("x1: " + x1);
        Object x = t.findLarger(123,"456");
        double x2 = t.findLarger(123, new Double(456)); // Line 5
    }
}

Dear friends,
Why does the variable x2 (Line 5) produce compiler error?
What exactly T is? Is it equivalent to E?

Kindly provide your help.
Regards,
Vijay

[Collins Mbianda]

Hi !!!

double x2 = t.findLarger(123, new Double(456));

is not compiling because the method is waiting for two arguments of the same type. You give him an int and a Double.
Note that integer litteral are considered like int. Thus 123 will be boxed as an Integer.

You should write this:

double x2 = t.findLarger(123D, new Double(456)); 

[Arek Orzechowski]

Hi,

I agree it should be

double x2 = t.findLarger(123d, new Double(456));

but I think the reason the original line is not compiling are not the parameters (123 is boxed as Integer) but the result, which is of class Number (common superclass of Integer and Double) and cannot be converted to double

this compiles ok:

Number x2 = t.findLarger(123, new Double(456));

[Collins Mbianda]

Originally posted by Arek Orzechowski:
Hi,
....
this compiles ok:

Number x2 = t.findLarger(123, new Double(456));

I agree with you that the code will compile, because Number is superclass of Integer and Double. Since we are talking of compilation it's ok.

I will just add that the code will throw a ClassCastException if you don't turned 123 into a double even if it compiles.

[Vijay Chandran]

Thanks for the reply!!!




Regards,
Vijay