[c#] Remove trailing zeros

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 tried with String.Format, but it returns 2.0000 and setting it to N0 rounds the other values as well.

This question is related to c# .net decimal

The answer is


This will work:

decimal source = 2.4200m;
string output = ((double)source).ToString();

Or if your initial value is string:

string source = "2.4200";
string output = double.Parse(source).ToString();

Pay attention to this comment.


Use the hash (#) symbol to only display trailing 0's when necessary. See the tests below.

decimal num1 = 13.1534545765;
decimal num2 = 49.100145;
decimal num3 = 30.000235;

num1.ToString("0.##");       //13.15%
num2.ToString("0.##");       //49.1%
num3.ToString("0.##");       //30%

I ran into the same problem but in a case where I do not have control of the output to string, which was taken care of by a library. After looking into details in the implementation of the Decimal type (see http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx), I came up with a neat trick (here as an extension method):

public static decimal Normalize(this decimal value)
{
    return value/1.000000000000000000000000000000000m;
}

The exponent part of the decimal is reduced to just what is needed. Calling ToString() on the output decimal will write the number without any trailing 0. E.g.

1.200m.Normalize().ToString();

how about this:

public static string TrimEnd(this decimal d)
    {
        string str = d.ToString();
        if (str.IndexOf(".") > 0)
        {
            str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "0+?$", " ");
            str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "[.]$", " ");
        }
        return str;
    }

Very simple answer is to use TrimEnd(). Here is the result,

double value = 1.00;
string output = value.ToString().TrimEnd('0');

Output is 1 If my value is 1.01 then my output will be 1.01


You can just set as:

decimal decNumber = 23.45600000m;
Console.WriteLine(decNumber.ToString("0.##"));

I found an elegant solution from http://dobrzanski.net/2009/05/14/c-decimaltostring-and-how-to-get-rid-of-trailing-zeros/

Basically

decimal v=2.4200M;

v.ToString("#.######"); // Will return 2.42. The number of # is how many decimal digits you support.

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

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

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
=>

This is simple.

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

Tested.

Cheers :)


The following code could be used to not use the string type:

int decimalResult = 789.500
while (decimalResult>0 && decimalResult % 10 == 0)
{
    decimalResult = decimalResult / 10;
}
return decimalResult;

Returns 789.5


try this code:

string value = "100";
value = value.Contains(".") ? value.TrimStart('0').TrimEnd('0').TrimEnd('.') : value.TrimStart('0');

I use this code to avoid "G29" scientific notation:

public static string DecimalToString(this decimal dec)
{
    string strdec = dec.ToString(CultureInfo.InvariantCulture);
    return strdec.Contains(".") ? strdec.TrimEnd('0').TrimEnd('.') : strdec;
}

EDIT: using system CultureInfo.NumberFormat.NumberDecimalSeparator :

public static string DecimalToString(this decimal dec)
{
    string sep = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator;
    string strdec = dec.ToString(CultureInfo.CurrentCulture);
    return strdec.Contains(sep) ? strdec.TrimEnd('0').TrimEnd(sep.ToCharArray()) : strdec;
}

try like this

string s = "2.4200";

s = s.TrimStart('0').TrimEnd('0', '.');

and then convert that to float


In my opinion its safer to use Custom Numeric Format Strings.

decimal d = 0.00000000000010000000000m;
string custom = d.ToString("0.#########################");
// gives: 0,0000000000001
string general = d.ToString("G29");
// gives: 1E-13

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"];


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);
}

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