Remove trailing zeros

Question

I have some fields returned by a collection as

2.4200
2.0044
2.0000

I want results like

2.42
2.0044
2

I t

Answer 1

Is it not as simple as this, if the input IS a string? You can use one of these:

string.Format("{0:G29}", decimal.Parse("2.0044"))

decimal.Parse("2.0044").ToString("G29")

2.0m.ToString("G29")

This should work for all input.

Update Check out the Standard Numeric Formats I've had to explicitly set the precision specifier to 29 as the docs clearly state:

However, if the number is a Decimal and the precision specifier is omitted, fixed-point notation is always used and trailing zeros are preserved

Update Konrad pointed out in the comments:

Watch out for values like 0.000001. G29 format will present them in the shortest possible way so it will switch to the exponential notation. string.Format("{0:G29}", decimal.Parse("0.00000001",System.Globalization.CultureInfo.GetCultureInfo("en-US"))) will give "1E-08" as the result.

Answer 2

This is simple.

decimal decNumber = Convert.ToDecimal(value);
        return decNumber.ToString("0.####");

Tested.

Cheers :)

Answer 3

Trying to do more friendly solution of DecimalToString (https://stackoverflow.com/a/34486763/3852139):

private static decimal Trim(this decimal value)
{
    var s = value.ToString(CultureInfo.InvariantCulture);
    return s.Contains(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator)
        ? Decimal.Parse(s.TrimEnd('0'), CultureInfo.InvariantCulture)
        : value;
}

private static decimal? Trim(this decimal? value)
{
    return value.HasValue ? (decimal?) value.Value.Trim() : null;
}

private static void Main(string[] args)
{
    Console.WriteLine("=>{0}", 1.0000m.Trim());
    Console.WriteLine("=>{0}", 1.000000023000m.Trim());
    Console.WriteLine("=>{0}", ((decimal?) 1.000000023000m).Trim());
    Console.WriteLine("=>{0}", ((decimal?) null).Trim());
}

Output:

=>1
=>1.000000023
=>1.000000023
=>

Answer 4

In case you want to keep decimal number, try following example:

number = Math.Floor(number * 100000000) / 100000000;

Answer 5

A very low level approach, but I belive this would be the most performant way by only using fast integer calculations (and no slow string parsing and culture sensitive methods):

public static decimal Normalize(this decimal d)
{
    int[] bits = decimal.GetBits(d);

    int sign = bits[3] & (1 << 31);
    int exp = (bits[3] >> 16) & 0x1f;

    uint a = (uint)bits[2]; // Top bits
    uint b = (uint)bits[1]; // Middle bits
    uint c = (uint)bits[0]; // Bottom bits

    while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
    {
        uint r;
        a = DivideBy10((uint)0, a, out r);
        b = DivideBy10(r, b, out r);
        c = DivideBy10(r, c, out r);
        exp--;
    }

    bits[0] = (int)c;
    bits[1] = (int)b;
    bits[2] = (int)a;
    bits[3] = (exp << 16) | sign;
    return new decimal(bits);
}

private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
    ulong total = highBits;
    total <<= 32;
    total = total | (ulong)lowBits;

    remainder = (uint)(total % 10L);
    return (uint)(total / 10L);
}

Answer 6

Depends on what your number represents and how you want to manage the values: is it a currency, do you need rounding or truncation, do you need this rounding only for display?

If for display consider formatting the numbers are x.ToString("")

http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx and

http://msdn.microsoft.com/en-us/library/0c899ak8.aspx

If it is just rounding, use Math.Round overload that requires a MidPointRounding overload

http://msdn.microsoft.com/en-us/library/ms131274.aspx)

If you get your value from a database consider casting instead of conversion: double value = (decimal)myRecord["columnName"];