Why Java's Math.round() cannot handle this number?

Question

I have code like this:

System.out.println("Math.round(1423562400L) => " + Math.round(1423562400L));

And the result is this:

Math.round(1423562400L) => 1423562368

This value is a 64-bit integer but it easily fits inside a 32-bit integer which should fit in a double. What is going on?

Answer 1

According to the JLS,

15.12.2.5. Choosing the Most Specific Method

If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run-time method dispatch. The Java programming language uses the rule that the most specific method is chosen.

The informal intuition is that one method is more specific than another if any invocation handled by the first method could be passed on to the other one without a compile-time type error.

It follows that Math.round(float) is more specific than Math.round(double), and since they're both valid overloads, the compiler chooses the former, resulting in lost data.

To correct your output, simply change 1423562400L to 1423562400d.