Truncate Two decimal places without rounding

Lets say I have a value of 3.4679 and want 3.46, how can I truncate to two decimal places that without rounding up?

I have tried the following but all three give me 3.47:

void Main()
{
    Console.Write(Math.Round(3.4679, 2,MidpointRounding.ToEven));
    Console.Write(Math.Round(3.4679, 2,MidpointRounding.AwayFromZero));
    Console.Write(Math.Round(3.4679, 2));
}

This returns 3.46, but just seems dirty some how:

void Main()
{
    Console.Write(Math.Round(3.46799999999 -.005 , 2));
}

Solution 1:

value = Math.Truncate(100 * value) / 100;

Beware that fractions like these cannot be accurately represented in floating point.

Solution 2:

It would be more useful to have a full function for real-world usage of truncating a decimal in C#. This could be converted to a Decimal extension method pretty easy if you wanted:

public decimal TruncateDecimal(decimal value, int precision)
{
    decimal step = (decimal)Math.Pow(10, precision);
    decimal tmp = Math.Truncate(step * value);
    return tmp / step;
}

If you need VB.NET try this:

Function TruncateDecimal(value As Decimal, precision As Integer) As Decimal
    Dim stepper As Decimal = Math.Pow(10, precision)
    Dim tmp As Decimal = Math.Truncate(stepper * value)
    Return tmp / stepper
End Function

Then use it like so:

decimal result = TruncateDecimal(0.275, 2);

or

Dim result As Decimal = TruncateDecimal(0.275, 2)

Solution 3:

Universal and fast method (without Math.Pow() / multiplication) for System.Decimal:

decimal Truncate(decimal d, byte decimals)
{
    decimal r = Math.Round(d, decimals);

    if (d > 0 && r > d)
    {
        return r - new decimal(1, 0, 0, false, decimals);
    }
    else if (d < 0 && r < d)
    {
        return r + new decimal(1, 0, 0, false, decimals);
    }

    return r;
}