Math.round bug - what to do?

Question

Math.Round(8.075, 2, MidpointRounding.AwayFromZero) returns 8.07, though it should return 8.08. Mysteriously enough, 7.075 is working fine, but 9.075 also returns 9.07!

What to do? Does anybody know a rounding method without such bugs?

Answer 1

If you count with 10 fingers, like humans do, you don't have any trouble expressing the decimal value 8.075 precisely:

  8.075 = 8 x 10^1 + 0 x 10^0 + 7 x 10^-1 + 5 x 10^-2

But computers count with 2 fingers, they need to express that value in powers of 2:

  8.075 = 1 x 2^3  + 0 x 2^2  + 0 x 2^1  + 0 x 2^0  + 0 x 2^-1 + 0 x 2^-2 + 0 x 2^-3 + 
          1 x 2^-4 + 0 x 2^-5 + 0 x 2^-6 + 1 x 2^-7 + 1 x 2^-8 + 0 x 2^-9 + 0 x 2^-10 +
          1 x 2^-11 + ...

I gave up with finger cramp typing the terms but the point is that no matter how many powers of 2 you add, you'll never get exactly 8.075m. A similar problem to how humans can never write the result of 10 / 3 precisely, it has an infinite number of digits in the fraction. You can only write the result of that expression accurately when you count with 6 fingers.

A processor of course doesn't have enough storage to store an infinite number of bits to represent a value. So they must truncate the digit sequence, a value of type double can store 53 bits.

As a result, the decimal value 8.075 gets rounded when it is stored in the processor. The sequence of 53 bits, converted back to decimal, is the value ~8.074999999999999289. Which then, as expected, gets rounded to 8.07 by your code.

If you want 10 finger math results, you'll need to use a data type that stores numbers in base 10. That's the System.Decimal type in .NET. Fix:

decimal result = Math.Round(8.075m, 2, MidpointRounding.AwayFromZero)

Note the usage of the letter m in the 8.075m literal in the snippet, a literal of type decimal. Which selects the Math.Round() overload that counts with 10 fingers, previously you used the overload that uses System.Double, the 2 finger version.

Do note that there's a significant disadvantage to calculating with System.Decimal, it is slow. Much, much slower than calculating with System.Double, a value type that's directly supported by the processor. Decimal math is done in software and is not hardware accelerated.

Answer 2

I am not a .net specialist, but these numbers can't be exactly represented as double, so the rounding is accurate if you take into account the real value of those 3 numbers:

7.075 ==> 7.07500000000000017763568394002504646778106689453125
8.075 ==> 8.074999999999999289457264239899814128875732421875
9.075 ==> 9.074999999999999289457264239899814128875732421875

More about floating-point precision: What Every Computer Scientist Should Know About Floating-Point Arithmetic.