According to MSDN, Math.Round will return the following: Math.Round(4.4); //Returns 4.0. Math.Round(4.5); //Returns 4.0. Math.Round(4.6); //Returns 5.0.
So, in my case doing Math.Round(objMyObject.Amount * 100.0) will not do the trick always. For example: Math.Round(123.454 * 100.0); // returns 12345 Math.Round(123.455 * 100.0); // returns 12345 Math.Round(123.456 * 100.0); // returns 12346 So I suppose I need to execute this: Math.Round((objMyObject.Amount * 100.0) - 0.5) to be sure that I get what I want. Am I missing something? -----Original Message----- From: Discussion of advanced .NET topics. [mailto:[EMAIL PROTECTED] Behalf Of Peter Ritchie Sent: Monday, August 25, 2008 12:48 PM To: ADVANCED-DOTNET@DISCUSS.DEVELOP.COM Subject: Re: [ADVANCED-DOTNET] Converting doubles into integers without rounding errors As John has alluded to, what you're seeing is a difference in rounding between formatting a double value as text and multiplying by 100 and dropping the decimals. If you have specific rounding logic, perform it before you convert to int. If you want the same value that you'd see as formatted text, convert to text first. I would recommend something like this: double tempAmount = objMyObject.Amount * 100.0; uint nIntAmount = Math.Round(tempAmount); =================================== This list is hosted by DevelopMentor® http://www.develop.com View archives and manage your subscription(s) at http://discuss.develop.com =================================== This list is hosted by DevelopMentor® http://www.develop.com View archives and manage your subscription(s) at http://discuss.develop.com
