How do I round down a decimal to 2 decimal places in .Net?

Question

This should be easy, but I can’t find a built in method for it, the .net framework must have a method to do this!

private decimal RoundDownTo2DecimalPlaces(decimal input)
{
   if (input < 0)
   {
      throw new Exception("not tested with negitive numbers");
   }

   // There must be a better way!
   return Math.Truncate(input * 100) / 100;
}

Do you want to round down the decimal itself or a text representation? — Henk Holterman, Sep 26 '10 at 10:13
@Henk, I need to output the value as text, but would rather do the rounding before the output. It is likly the value will have to feed into other calcs later. — Ian Ringrose, Sep 26 '10 at 10:17
possible duplicate of [Round a double to 2 significant figures after decimal point](http://stackoverflow.com/questions/2808535/round-a-double-to-2-significant-figures-after-decimal-point) — ecatmur, Sep 17 '12 at 17:24
Did you really accept your own post as an answer when it doesn't even have code? lol — codemonkeyliketab, Apr 22 '16 at 19:45

FinnNk · Answer 1 · 2010-09-26T10:40:43.200

21

If you are rounding down then you need:

Math.Floor(number * 100) / 100;

if you are looking for something called 'bankers rounding' (probably not if it's for output and not for statistics/summing) then:

Math.Round(number, 2);

Finally if you want, not sure what the correct term is, 'normal rounding':

Math.Round(number, 2, MidpointRounding.AwayFromZero);

edited Sep 26 '10 at 10:40

answered Sep 26 '10 at 10:34

FinnNk

3,229
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Look at my response below: MidpointRounding.AwayFromZero or MidpointRounding.ToEven specify how to handle numbers ending with '5': the MidpointRounding.ToEven specify that 1.135 should be round to 1.13, and 1.145 to 1.15 the MidpointRounding.AwayFromZero specify that 1.135 should be round to 1.14, and 1.145 to 1.15 – Andrea Parodi Sep 26 '10 at 10:56

Andrea Parodi · Answer 2 · 2010-09-26T11:09:02.187

Use Math.Floor if you want to round down tha value, or Math.Round if you want to get an exact round. Math.Truncate simply remove the decimal part of the number,so you get bad results for negative numbers:

var result= Math.Floor(number * 100) / 100;

Math.Floor always return the smallest integral value that is lesser (Floor ) or greater (Ceiling) than the specified value. So you don't get a correct rounding. Example:

Math.Floor(1.127 * 100) / 100 == 1.12 //should be 1.13 for an exact round
Math.Ceiling(1.121 * 100) / 100 == 1.13 //should be 1.12 for an exact round

Always prefer the version of Math.Round containing the mid-point rounding param. This param specify how to handle mid-point values (5) as last digit.

If you don't specify AwayFromZero as the value for param, you'll get the default behaviour, which is ToEven. For example, using ToEven as rounding method, you get:

Math.Round(2.025,2)==2.02 
Math.Round(2.035,2)==2.04

instead, using MidPoint.AwayFromZero param:

Math.Round(2.025,2,MidpointRounding.AwayFromZero)==2.03
Math.Round(2.035,2,MidpointRounding.AwayFromZero)==2.04

So, for a normal rounding, it's best to use this code:

var value=2.346;
var result = Math.Round(value, 2, MidpointRounding.AwayFromZero);

score 3 · Answer 3 · answered Sep 26 '10 at 10:16

3

Math.Floor(number * 100) / 100;

answered Sep 26 '10 at 10:16

Itay Karo

17,370
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butter then my solution, as Math.Floor give the correct result with negitive numbers. see http://stackoverflow.com/questions/14/whats-the-difference-between-math-floor-and-math-truncate-in-net – Ian Ringrose Sep 26 '10 at 10:27

KMån · Answer 4 · 2010-09-26T10:28:22.863

Use .Truncate() to get exact amount, or .Round() to round off.

decimal dNum = (decimal)165.6598F;
decimal dTruncated = (decimal)(Math.Truncate((double)dNum*100.0) / 100.0); //Will give 165.65
decimal dRounded = (decimal)(Math.Round((double)dNum, 2)); //Will give 165.66

Or you can make an extension method to run it like dNum.ToTwoDecimalPlaces();

public static class Extensions
{ 
    public static decimal ToTwoDecimalPlaces(this decimal dNum)
    {
        return ((decimal)(Math.Truncate((double)dNum*100.0) / 100.0));
    }
}

score 0 · Accepted Answer · answered Sep 30 '10 at 09:32

0

There is no build in method in the .net framework to do this, other answers say how to write your own code.

answered Sep 30 '10 at 09:32

Ian Ringrose

50,487
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How do I round down a decimal to 2 decimal places in .Net?

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