Cast object (type double) to int

Question

Okay, so if I have this code

double a=1.5;
int b=(int)a;
System.out.println(b);

Everything works fine, but

Object a=1.5;
int b=(int)a;
System.out.println(b);

gives the following error after running (Eclipse doesn't give any error)

java.lang.ClassCastException: java.lang.Double cannot be cast to java.lang.Integer

Though, when I do

Object a=1.5;
double b=(double)a;
int c=(int)b;
System.out.println(c);

or

Object a=1.5;
int b=(int)(double)a;
System.out.println(b);

Nothing's wrong again.

Why do you have to cast it to double first ?

Answer 1

When you declare the object Object a = 1.5 you can tell by checking System.out.println(a.getClass()) that the object is in fact cast to a Double instance. This can again be cast to a double because of unboxing conventions. After that the double value can be cast to an int.

There are however no unboxing conventions to cast from a Double instance to an int, so the runtime will issue an ClassCastException if you try and do that. It cannot directly go from Double to Integer.

Answer 2

When you're casting from Object, you're unboxing from the wrapper type... and you can only unbox to the original type, basically. It's effectively a cast to the relevant wrapper type, followed by a call to the appropriate xxxValue method. So this:

Object x = ...;
double d = (double) x;

is equivalent to:

Object x = ...;
double d = ((Double) x).doubleValue();

That cast to Double will obviously fail if x isn't a reference to a Double.

So your problematic code is equivalent to:

Object a = Double.valueOf(1.5); // Auto-boxing in the original code
int b = ((Integer) a).intValue(); // Unboxing in the original code
System.out.println(b);

Hopefully now it's obvious why that would fail - because the first line creates a Double which you're then trying to cast to Integer.