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Asked: 2026-05-30T14:05:29+00:00

Possible Duplicate: Why is floating point arithmetic in C# imprecise? If I loop through

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Possible Duplicate:
Why is floating point arithmetic in C# imprecise?

If I loop through numerous random doubles, and “round” them to two fractional digit places, each individual round appears to be correct (0.02, 0.01, 0.00, etc).

However, there appears to be a very small fractional part that is kept along with the round.

double total = 0;

for (int i = 0; i < 10000; i++)
{
    total += Math.Round(random.NextDouble() * 0.02, 2);
}

Console.WriteLine(total);

Sample Outputs:

100.600000000006

99.7400000000059

Anyone care to explain why this happens is a more intuitive way?

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1 Answer

  1. Editorial Team

    System.Double and System.Float are base 2 floating point types. There are many finite decimal values that have an infinite representation in base 2, much as 1/3 has an infinite representation in base 10. Therefore, when you round to such a value, the binary representation is approximate. To avoid this problem, use the decimal type, which is a base 10 floating point type.

    There must be 100 duplicates of this question on stackoverflow, but I am on my phone, which makes it inconvenient to find them and link to them.

    For more information, look at the wikipedia article for IEEE double.

    Many will say that doubles are “not exact”, which is false. Every double value represents an exact value that can be represented exactly in base 10 (except NaN and infinity, of course). That’s because 2 is one of the prime factors of 10. The only approximation is when you try to represent certain decimal fractions (or other rational numbers whose denominator has at least one prime factor other than 2).

    The best way to understand this, for me at least, is to work out, on paper, the binary representations of a few fractions. For example, try 0.5, 0.625, 3.25, 5/16, 1/3, 0.2, and 0.3.

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