rounded to the nearest 10's place in c#

Question

I want to The numbers are being rounded to the nearest 10's place. For example, a number like 17.3 is being rounded to 20.0. and want to be allow three significant digits. Meaning for round to the nearest tenth of a percent as the last step of the process.

Sample :

    the number is 17.3 ,i want round to 20 ,

    and this number is 13.3 , i want round to 10 ?

How can i do this ?

score 7 · Answer 1 · edited May 23 '17 at 12:12

Chris Charabaruk gives you your desired answer here

To get to the core, here is his solution as an extension method:

public static class ExtensionMethods
{
    public static int RoundOff (this int i)
    {
        return ((int)Math.Round(i / 10.0)) * 10;
    }
}

int roundedNumber = 236.RoundOff(); // returns 240
int roundedNumber2 = 11.RoundOff(); // returns 10

//edit: This method only works for int values. You'd have to edit this method to your liking. f.e.: public static class ExtensionMethods

{
    public static double RoundOff (this double i)
    {
       return (Math.Round(i / 10.0)) * 10;
    }
}

/edit2: As corak stated you should/could use

Math.Round(value / 10, MidpointRounding.AwayFromZero) * 10

Is the cast to `double` really necessary? And to avoid bankers rounding shenanigans, I'd use `Math.Round(value / 10, MidpointRounding.AwayFromZero) * 10` — Corak, Aug 21 '13 at 06:19

score 2 · Answer 2 · answered Aug 21 '13 at 06:19

2

Other answers are correct also, but here's how you'd do it without Math.Round:

((int)((17.3 + 5) / 10)) * 10 // = 20
((int)((13.3 + 5) / 10)) * 10 // = 10
((int)((15.0 + 5) / 10)) * 10 // = 20

answered Aug 21 '13 at 06:19

Eren Ersönmez

38,383
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What specific reason do you have for doing it without `Round`? Are you assuming you're doing it more efficiently than the math library is? – arkon Oct 17 '22 at 09:43

score 0 · Answer 3 · answered Aug 21 '13 at 06:16

0

Try this-

double d1 = 17.3;
int rounded1 = ((int)Math.Round(d/10.0)) * 10; // Output is 20

double d2 = 13.3;
int rounded2 = ((int)Math.Round(d/10.0)) * 10; // Output is 10

answered Aug 21 '13 at 06:16

Microsoft DN

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It will give output `20` if value is 15. What are you expecting? – Microsoft DN Aug 21 '13 at 06:21

score 0 · Answer 4 · answered Aug 21 '13 at 06:18

0

double Num = 16.6;
int intRoundNum = (Convert.ToInt32(Math.Round(Num / 10)) * 10);
Console.WriteLine(intRoundNum);

answered Aug 21 '13 at 06:18

Prabhakaran Parthipan

4,193
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score 0 · Answer 5 · answered Jan 15 '21 at 20:02

If you want to avoid casting or pulling in the math library you could also use the modulus operator and do something like the following:

int result = number - (number % 10);
if (number % 10 >= 5)
{
    result += 10;
}

For your given numbers:

number	processing	result
13.3	13.3 - (3.3)	10
17.3	17.3 - (7.3) + 10	20

rounded to the nearest 10's place in c#

5 Answers5

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