When displaying the value of a decimal currently with .ToString(), it's accurate to like 15 decimal places, and since I'm using it to represent dollars and cents, I only want the output to be 2 decimal places.
Do I use a variation of .ToString() for this?
decimalVar.ToString("#.##"); // returns ".5" when decimalVar == 0.5m or
decimalVar.ToString("0.##"); // returns "0.5" when decimalVar == 0.5m or
decimalVar.ToString("0.00"); // returns "0.50" when decimalVar == 0.5m 
Decimal and Double type ToString method accepts argument for formatting. Try converting your value to Decimal/Double first.
I know this is an old question, but I was surprised to see that no one seemed to post an answer that;
This is what I would use:
decimal.Round(yourValue, 2, MidpointRounding.AwayFromZero); 
decimalVar.ToString("F"); This will:
23.456 → 23.4623 → 23.00; 12.5 → 12.50Ideal for displaying currency.
Check out the documentation on ToString("F") (thanks to Jon Schneider).
. might be replaced with , based on the culture. You should pass along CultureInfo.InvariantCulture as the second argument to disable this.If you just need this for display use string.Format
String.Format("{0:0.00}", 123.4567m); // "123.46" http://www.csharp-examples.net/string-format-double/
The "m" is a decimal suffix. About the decimal suffix:
If you want it formatted with commas as well as a decimal point (but no currency symbol), such as 3,456,789.12...
decimalVar.ToString("n2"); Given decimal d=12.345; the expressions d.ToString("C") or String.Format("{0:C}", d) yield $12.35 - note that the current culture's currency settings including the symbol are used.
Note that "C" uses number of digits from current culture. You can always override default to force necessary precision with C{Precision specifier} like String.Format("{0:C2}", 5.123d).
There's a very important characteristic of Decimal that isn't obvious:
A
Decimal'knows' how many decimal places it has based upon where it came from
The following may be unexpected :
Decimal.Parse("25").ToString() => "25" Decimal.Parse("25.").ToString() => "25" Decimal.Parse("25.0").ToString() => "25.0" Decimal.Parse("25.0000").ToString() => "25.0000" 25m.ToString() => "25" 25.000m.ToString() => "25.000" Doing the same operations with Double will result in zero decimal places ("25") for all of the above examples.
If you want a decimal to 2 decimal places there's a high likelihood it's because it's currency in which case this is probably fine for 95% of the time:
Decimal.Parse("25.0").ToString("c") => "$25.00" Or in XAML you would use {Binding Price, StringFormat=c}
One case I ran into where I needed a decimal AS a decimal was when sending XML to Amazon's webservice. The service was complaining because a Decimal value (originally from SQL Server) was being sent as 25.1200 and rejected, (25.12 was the expected format).
All I needed to do was Decimal.Round(...) with 2 decimal places to fix the problem regardless of the source of the value.
// generated code by XSD.exe StandardPrice = new OverrideCurrencyAmount() { TypedValue = Decimal.Round(product.StandardPrice, 2), currency = "USD" } TypedValue is of type Decimal so I couldn't just do ToString("N2") and needed to round it and keep it as a decimal.
Here is a little Linqpad program to show different formats:
void Main() { FormatDecimal(2345.94742M); FormatDecimal(43M); FormatDecimal(0M); FormatDecimal(0.007M); } public void FormatDecimal(decimal val) { Console.WriteLine("ToString: {0}", val); Console.WriteLine("c: {0:c}", val); Console.WriteLine("0.00: {0:0.00}", val); Console.WriteLine("0.##: {0:0.##}", val); Console.WriteLine("==================="); } Here are the results:
ToString: 2345.94742 c: $2,345.95 0.00: 2345.95 0.##: 2345.95 =================== ToString: 43 c: $43.00 0.00: 43.00 0.##: 43 =================== ToString: 0 c: $0.00 0.00: 0.00 0.##: 0 =================== ToString: 0.007 c: $0.01 0.00: 0.01 0.##: 0.01 =================== 
Very rarely would you want an empty string if the value is 0.
decimal test = 5.00; test.ToString("0.00"); //"5.00" decimal? test2 = 5.05; test2.ToString("0.00"); //"5.05" decimal? test3 = 0; test3.ToString("0.00"); //"0.00" The top rated answer is incorrect and has wasted 10 minutes of (most) people's time.
"#" means digit of number (if necessary) (without padding if not needed) "0" means digit of number (no mater what) (padded with zeros if not available)Mike M.'s answer was perfect for me on .NET, but .NET Core doesn't have a decimal.Round method at the time of writing.
In .NET Core, I had to use:
decimal roundedValue = Math.Round(rawNumber, 2, MidpointRounding.AwayFromZero); A hacky method, including conversion to string, is:
public string FormatTo2Dp(decimal myNumber) { // Use schoolboy rounding, not bankers. myNumber = Math.Round(myNumber, 2, MidpointRounding.AwayFromZero); return string.Format("{0:0.00}", myNumber); } None of these did exactly what I needed, to force 2 d.p. and round up as 0.005 -> 0.01
Forcing 2 d.p. requires increasing the precision by 2 d.p. to ensure we have at least 2 d.p.
then rounding to ensure we do not have more than 2 d.p.
Math.Round(exactResult * 1.00m, 2, MidpointRounding.AwayFromZero) 6.665m.ToString() -> "6.67" 6.6m.ToString() -> "6.60" The top-rated answer describes a method for formatting the string representation of the decimal value, and it works.
However, if you actually want to change the precision saved to the actual value, you need to write something like the following:
public static class PrecisionHelper { public static decimal TwoDecimalPlaces(this decimal value) { // These first lines eliminate all digits past two places. var timesHundred = (int) (value * 100); var removeZeroes = timesHundred / 100m; // In this implementation, I don't want to alter the underlying // value. As such, if it needs greater precision to stay unaltered, // I return it. if (removeZeroes != value) return value; // Addition and subtraction can reliably change precision. // For two decimal values A and B, (A + B) will have at least as // many digits past the decimal point as A or B. return removeZeroes + 0.01m - 0.01m; } } An example unit test:
[Test] public void PrecisionExampleUnitTest() { decimal a = 500m; decimal b = 99.99m; decimal c = 123.4m; decimal d = 10101.1000000m; decimal e = 908.7650m Assert.That(a.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture), Is.EqualTo("500.00")); Assert.That(b.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture), Is.EqualTo("99.99")); Assert.That(c.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture), Is.EqualTo("123.40")); Assert.That(d.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture), Is.EqualTo("10101.10")); // In this particular implementation, values that can't be expressed in // two decimal places are unaltered, so this remains as-is. Assert.That(e.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture), Is.EqualTo("908.7650")); } The most applicable solution is
decimalVar.ToString("#.##"); 
You can use system.globalization to format a number in any required format.
For example:
system.globalization.cultureinfo ci = new system.globalization.cultureinfo("en-ca"); If you have a decimal d = 1.2300000 and you need to trim it to 2 decimal places then it can be printed like this d.Tostring("F2",ci); where F2 is string formating to 2 decimal places and ci is the locale or cultureinfo.
for more info check this link
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
https://msdn.microsoft.com/en-us/library/dwhawy9k%28v=vs.110%29.aspx
This link explains in detail how you can handle your problem and what you can do if you want to learn more. For simplicity purposes, what you want to do is
double whateverYouWantToChange = whateverYouWantToChange.ToString("F2"); if you want this for a currency, you can make it easier by typing "C2" instead of "F2"
Double Amount = 0; string amount; amount=string.Format("{0:F2}", Decimal.Parse(Amount.ToString())); If you need to keep only 2 decimal places (i.e. cut off all the rest of decimal digits):
decimal val = 3.14789m; decimal result = Math.Floor(val * 100) / 100; // result = 3.14 If you need to keep only 3 decimal places:
decimal val = 3.14789m; decimal result = Math.Floor(val * 1000) / 1000; // result = 3.147 
var arr = new List<int>() { -4, 3, -9, 0, 4, 1 }; decimal result1 = arr.Where(p => p > 0).Count(); var responseResult1 = result1 / arr.Count(); decimal result2 = arr.Where(p => p < 0).Count(); var responseResult2 = result2 / arr.Count(); decimal result3 = arr.Where(p => p == 0).Count(); var responseResult3 = result3 / arr.Count(); Console.WriteLine(String.Format("{0:#,0.000}", responseResult1)); Console.WriteLine(String.Format("{0:#,0.0000}", responseResult2)); Console.WriteLine(String.Format("{0:#,0.00000}", responseResult3)); you can put as many 0 as you want.
