Official Javadoc says that Math.floor()
returns a double
that is \"equal to a mathematical integer\", but then why shouldn\'t it return an in
It's for precision. The double data-type has a 53 bit mantissa. Among other things that means that a double can represent all whole up to 2^53 without precision loss.
If you store such a large number in an integer you will get an overflow. Integers only have 32 bits.
Returning the integer as a double is the right thing to do here because it offers a much wider usefull number-range than a integer could.