Java: Inaccuracy using double [duplicate]

Answer 1

From The Floating-Point Guide:

Because internally, computers use a format (binary floating-point) that cannot accurately represent a number like 0.1, 0.2 or 0.3 at all.

When the code is compiled or interpreted, your “0.1” is already rounded to the nearest number in that format, which results in a small rounding error even before the calculation happens.

That accounts for your first example. The second one only involves integers, not fractions, and integers can be represented exactly in the binary floating-point format (up to 52 bits).

Answer 2

doubles are not exact.

It is because there are infinite possible real numbers and only finite number of bits to represent these numbers.

have a look at: what every programmer should know about floating point arithmetic

Answer 3

Some decimals cannot be exactly represented by double values. 0.3 is one of those values.

All integer values less than a certain number (I forget which) happen to have an exact representation by a double value, so you don't see the approximation.

Consider how we think of numbers: the number 123 is represented as (1 * 100) + (2 * 10) + (3 * 1). We use 10 as our base. Binary numbers use two. So when you look at fractions of a number, how could you represent 0.3 by adding individual powers of 2? You can't. The best you can come up with is about 0.30000000000000004 (I'd have to see the exact binary digits to see how it reaches that).

Answer 4

They're getting rounded in the second case. See also Ruby - Multiplication issue — it's the same problem.