Rounding of nearest 0.5

Question

I want to be rounded off this way

13.1, round to 13.5
13.2, round to 13.5
13.3, round to 13.5
13.4, round to 13.5
13.5 = 13.5
13.6, round to 14.0
13.7, roun         


        

Answer 1

This works, I just tested it;

double a = 13.3;
var rn  =  a % 0.5 == 0 ? 1 : 0;
Math.Round(a, rn);

Answer 2

A simple way of doing this without the builtin method of c# (if you want ) its writen i c++ (I once lacked the round function in c++ ) but you can easly change it to c# syntax

int round(float nNumToRound)
{

// Variable definition
int nResult;

// Check if the number is negetive
if (nNumToRound > 0)
{
    // its positive, use floor.
    nResult = floor(nNumToRound + 0.5);
}
else if (nNumToRound < 0)
{
    // its negtive, use ceil 
    nResult = ceil(nNumToRound - 0.5);
}

return (nResult);

}

Answer 3

Nearest 0.5 for 13.6 and 13.7 is 13.5, so you have correct solution.

for yours table of values:

var value = 13.5;
var reminder = value % (int)value;
var isMiddle = Math.Abs(reminder - 0.5) < 0.001;
var result =  (isMiddle ? Math.Round(value * 2, MidpointRounding.AwayFromZero): Math.Round(value)*2)/ 2;

Answer 4

var a = d == (((int)d)+0.5) ? d : Math.Round(d);

d is a double.

Answer 5

If it is required for 13.1, round to 13.5 and 13.9, round to 14.0, then:

double a = 13.1;
double rounded = Math.Ceil(a * 2) / 2;

Answer 6

num = (num % 0.5 == 0 ? num : Math.Round(num));

works well for you solution heres the complete console program

static void Main(string[] args)
        {
            double[] a = new double[]{
                13.1,13.2,13.3D,13.4,13.5,13.6,13.7,13.8,13.9,13.58,13.49,13.55,
            };
            foreach (var b in a)
            {
                Console.WriteLine("{0}-{1}",b,b % 0.5 == 0 ? b : Math.Round(b));
            }
            Console.ReadKey();
        }

you simply would need to change 0.5 to some other number if the rounding requirement changes in future