C# Math.Round Error?

Ok, I realize it's early on a Sunday so I hope I'm just missing something obvious:

I have this function:

private decimal CashConversion(decimal amount, decimal ratio)
{
    if (ratio == 1) return amount;

    decimal convertedAmount = amount / ratio;
    return Math.Round(convertedAmount, 2);
}

When I call it like this:

decimal ta开发者_如何转开发x = CashConversion(96.53, 15);

The "tax" variable is equal to 6.43. However, 96.53/15 is 6.435333333333333. Rounding that to 2 places should return 6.44. Am I missing something here?

Check the documentation for Math.Round: As 2 is even, and the next digit after the second is 5, the value is rounded down, according to the IEEE Standard 754, section 4. It's called banker's rounding.

It's not an error, but an expected behaviour. Maybe not the one you were expecting though.

If you want the "mathematically correct" behaviour though, you can call the Decimal.Round(Decimal, Int32, MidpointRounding) overload, as in:

Math.Round(convertedAmount, 2, MidpointRounding.AwayFromZero);

That's indeed what is expected: 6.435 would round to 6.44:

When d is exactly halfway between two rounded values, the result is the rounded value that has an even digit in the [(decimals+1)th decimal position]. For example, when rounded to two decimals, the value 2.345 becomes 2.34 and the value 2.355 becomes 2.36[, and 2.3653333 becomes 2.37]. This process is known as rounding toward even, or rounding to nearest.

By default, Math.Round uses banker's rounding. You probably are expecting it to use midpoint rounding. To force this behavior, try this:

Math.Round(convertedAmount, 2, MidpointRounding.AwayFromZero);