Why does Math.Round(2.5) return 2 instead of 3?

Question

In C#, the result of Math.Round(2.5) is 2.

It is supposed to be 3, isn\'t it? Why is it 2 instead in C#?

Answer 1

Silverlight doesn't support the MidpointRounding option. Here's an extension method for Silverlight that adds the MidpointRounding enum:

public enum MidpointRounding
{
    ToEven,
    AwayFromZero
}

public static class DecimalExtensions
{
    public static decimal Round(this decimal d, MidpointRounding mode)
    {
        return d.Round(0, mode);
    }

    /// 

    /// Rounds using arithmetic (5 rounds up) symmetrical (up is away from zero) rounding
    /// 
    /// A Decimal number to be rounded.
    /// The number of significant fractional digits (precision) in the return value.
    /// The number nearest d with precision equal to decimals. If d is halfway between two numbers, then the nearest whole number away from zero is returned.
    public static decimal Round(this decimal d, int decimals, MidpointRounding mode)
    {
        if ( mode == MidpointRounding.ToEven )
        {
            return decimal.Round(d, decimals);
        }
        else
        {
            decimal factor = Convert.ToDecimal(Math.Pow(10, decimals));
            int sign = Math.Sign(d);
            return Decimal.Truncate(d * factor + 0.5m * sign) / factor;
        }
    }
}

Source: http://anderly.com/2009/08/08/silverlight-midpoint-rounding-solution/