Java Double variables have strange values [duplicate]

Answer 1

You probably should read "What Every Computer Scientist Should Know About Floating-Point Arithmetic".

In short, not all real values can be represented exactly with a double, so they should be avoided when exact answers are required.

Answer 2

In IEEE-754 binary double, we need to consider 1.1 and 1.2 in the binary representation:

1.2 = 0b1.001100110011001100110011001100110011001100110011001100110011...
1.1 = 0b1.000110011001100110011001100110011001100110011001100110011001...

note that we need infinitely many bits to represent them exactly in binary. double only has 53 bits of significance, we have to chop off the numbers:

1.2 = 0b1.001100110011001100110011001100110011001100110011001100110011...
1.1 = 0b1.000110011001100110011001100110011001100110011001100110011001...
                                                              ^ round from here
==>
1.2 ~ 0b1.0011001100110011001100110011001100110011001100110011
      (= exactly 1.1999999999999999555910790149937383830547332763671875)
1.1 ~ 0b1.0001100110011001100110011001100110011001100110011010
      (= exactly 1.100000000000000088817841970012523233890533447265625)

Hence 1.2 - 1.1 is:

  1.2 ~ 0b1.0011001100110011001100110011001100110011001100110011
- 1.1 ~ 0b1.0001100110011001100110011001100110011001100110011010
————————————————————————————————————————————————————————————————
        0b0.00011001100110011001100110011001100110011001100110010000
        (= exactly 0.09999999999999986677323704498121514916419982910156250000)

We can actually compute 60 / 0.0999999999999998667732370449812151491641998291015625 exactly, which gives

600.0000000000007993605777301137740672368493927467455286920109359612256820927...
                ^ 16th significant figure

that matches OP's result of

600.0000000000008

Answer 3

You've discovered the wonders of floating point arithmetic. Since the internal representation of a number is in binary, certain decimal numbers can't be represented precisely. As a result, some arithmetic operations will give unexpected results in the least significant digits.

This is covered in exhausting detail in, for example, numerical methods courses, but the Wikipedia article isn't bad.

Answer 4

This is not precisely a "rounding issue." Certain numbers cannot be exactly represented in binary floating point. Just like 1/3 cannot be represented exactly in decimal, since nobody has an infinite supply of paper. (or bits).

Answer 5

Because the double primitive is a floating point data type which trades precision for the ability to hold a larger range of values.

If you need arbitrary precision you should be using BigDecimal instead.

Answer 6

You should use BigDecimal to get correct result.