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;
}

Answer 1

Math.Floor(number * 100) / 100;

Answer 2

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));
    }
}

Answer 3

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);

Answer 4

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);

Answer 5

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