Rounding down to 2 decimal places in c#

Question

How can I multiply two decimals and round the result down to 2 decimal places?

For example if the equation is 41.75 x 0.1 the result will be 4.175. If I do this in c# with decimals it will automatically round up to 4.18. I would like to round down to 4.17.

I tried using Math.Floor but it just rounds down to 4.00. Here is an example:

Math.Floor (41.75 * 0.1);

Answer 1

The Math.Round(...) function has an Enum to tell it what rounding strategy to use. Unfortunately the two defined won't exactly fit your situation.

The two Midpoint Rounding modes are:

1. AwayFromZero - When a number is halfway between two others, it is rounded toward the nearest number that is away from zero. (Aka, round up)
2. ToEven - When a number is halfway between two others, it is rounded toward the nearest even number. (Will Favor .16 over .17, and .18 over .17)

What you want to use is Floor with some multiplication.

var output = Math.Floor((41.75 * 0.1) * 100) / 100;

The output variable should have 4.17 in it now.

In fact you can also write a function to take a variable length as well:

public decimal RoundDown(decimal i, double decimalPlaces)
{
   var power = Convert.ToDecimal(Math.Pow(10, decimalPlaces));
   return Math.Floor(i * power) / power;
}

Answer 2

public double RoundDown(double number, int decimalPlaces)
{
     return Math.Floor(number * Math.Pow(10, decimalPlaces)) / Math.Pow(10, decimalPlaces);
}

Answer 3

There is no native support for precision floor/ceillin in c#.

You can however mimic the functionality by multiplying the number, the floor, and then divide by the same multiplier.

eg,

decimal y = 4.314M;
decimal x = Math.Floor(y * 100) / 100;  // To two decimal places (use 1000 for 3 etc)
Console.WriteLine(x);  // 4.31

Not the ideal solution, but should work if the number is small.

Answer 4

从.NET Core 3.0和即将推出的.NET Framework 5.0 ，以下内容有效

Math.Round(41.75 * 0.1, 2, MidpointRounding.ToZero)

Answer 5

One more solution is to make rounding toward zero from rounding away from zero. It should be something like this:

    static decimal DecimalTowardZero(decimal value, int decimals)
    {
        // rounding away from zero
        var rounded = decimal.Round(value, decimals, MidpointRounding.AwayFromZero);

        // if the absolute rounded result is greater 
        // than the absolute source number we need to correct result
        if (Math.Abs(rounded) > Math.Abs(value))
        {
            return rounded - new decimal(1, 0, 0, value < 0, (byte)decimals);
        }
        else
        {
            return rounded;
        }
    }

Answer 6

This is my Float-Proof Round Down.

    public static class MyMath
{
    public static double RoundDown(double number, int decimalPlaces)
    {
        string pr = number.ToString();
        string[] parts = pr.Split('.');
        char[] decparts = parts[1].ToCharArray();
        parts[1] = "";
        for (int i = 0; i < decimalPlaces; i++)
        {
            parts[1] += decparts[i];
        }
        pr = string.Join(".", parts);
        return Convert.ToDouble(pr);
    }
}

Answer 7

I've found that the best method is to use strings; the binary vagaries of Math tend to get things wrong, otherwise. One waits for .Net 5.0 to make this fact obsolete. No decimal places is a special case: you can use Math.Floor for that. Otherwise, we ToString the number with one more decimal place than is required, then parse that without its last digit to get the answer:

/// <summary>
/// Truncates a Double to the given number of decimals without rounding
/// </summary>
/// <param name="D">The Double</param>
/// <param name="Precision">(optional) The number of Decimals</param>
/// <returns>The truncated number</returns>
public static double RoundDown(this double D, int Precision = 0)
{
  if (Precision <= 0) return Math.Floor(D);
  string S = D.ToString("0." + new string('0', Precision + 1));
  return double.Parse(S.Substring(0, S.Length - 1));
}

Answer 8

If you want to round down any double to specific decimal places, if doesn´t matter if is the midpoint, you can use:

    public double RoundDownDouble(double number, int decimaPlaces)
    {
        var tmp = Math.Pow(10, decimaPlaces);
        return Math.Truncate(number * tmp) / tmp;
    }