Why does Math.Round give me two different values?

Question

class Program {
    static void Main(string[] args) {
      double d = 120.5;
      Console.WriteLine(Math.Round(120.5)); //121
      Console.WriteLine(Math.Round(d)); // 120
    }
}

When a variable is passed as an argument into Math.Round it produces an answer similar to Convert.ToInt32 where floating numbers are rounded off to the nearest even number if the trailing tenth number is 0.5.

Anyone can kindly explain? Thanks in advance.

Thanks for the answers! I use Replit most of the time, that's the output I got. But seeing the replies, I tested it again in VS and I got both 120. I guess there's a bug in replit? Kindly refer to attachments.

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What .NET version are you using? I couldn't reproduce this on .NET Framework 4.7.2 or .NET 6.0. — 41686d6564 stands w. Palestine, Oct 03 '22 at 11:08
Possible duplicate https://stackoverflow.com/questions/72362928/does-math-round-in-net-round-certain-values-incorrectly — Ramji, Oct 03 '22 at 11:12
`120.5` is exactly representable in an IEEE float and thus is not subject to rounding problems, so I doubt very much that this is actually reproducible. I suspect if you run this yourself you'll get the same results. — Matthew Watson, Oct 03 '22 at 11:31

Hossein Sabziani · Accepted Answer · 2022-10-03T11:35:11.497

I tested your code and returns 120 in two modes. It is good to know that the Math.Round() has features that you can use.

For example, you can say to always round to a number that is further from zero:

 double d = 120.5;

 Console.WriteLine(Math.Round(d,MidpointRounding.AwayFromZero)); //always 121

or final digit is even:

 Console.WriteLine(Math.Round(d,MidpointRounding.ToEven)); //always 120

and other features like MidpointRounding.ToNegativeInfinity, MidpointRounding.ToPositiveInfinity ....

Why does Math.Round give me two different values?

1 Answers1