Rounding double values in C#

Question

I want a rounding method on double values in C#. It needs to be able to round a double value to any rounding precision value. My code on hand looks like:

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Answer 1

double d = 1.2345;

Math.Round(d, 2);

the code above should do the trick.

Answer 2

The examples using decimal casting provided in Jimmy's answer don't answer the question, since they do not show how to round a double value to any rounding precision value as requested. I believe the correct answer using decimal casting is the following:

    public static double RoundI(double number, double roundingInterval)
    {
        return (double)((decimal)roundingInterval * Math.Round((decimal)number / (decimal)roundingInterval, MidpointRounding.AwayFromZero));
    }

Because it uses decimal casting, this solution is subject to the casting errors mentioned by Jeppe Stig Nielsen in his comment to Jimmy's answer.

Also, note that I specified MidpointRounding.AwayFromZero, since that is consistent with the requester's specification that RoundI(1.2345, 0.001) should give 1.235.

Answer 3

If you actually need to use double just replace it below and it will work but with the usual precision problems of binary floating-point arithmetics.

There's most certainly a better way to implement the "rounding" (almost a kind of bankers' rounding) than my string juggling below.

public static decimal RoundI(decimal number, decimal roundingInterval)
{
   if (roundingInterval == 0) { return 0;}

   decimal intv = Math.Abs(roundingInterval);
   decimal modulo = number % intv;
   if ((intv - modulo) == modulo) {
       var temp = (number - modulo).ToString("#.##################");
       if (temp.Length != 0 && temp[temp.Length - 1] % 2 == 0) modulo *= -1;
   }
    else if ((intv - modulo) < modulo)
        modulo = (intv - modulo);
    else
        modulo *= -1;

    return number + modulo;
}

Answer 4

Example of using decimal, as Kibbee pointed out

double d = 1.275;
Math.Round(d, 2);          // 1.27
Math.Round((decimal)d, 2); // 1.28