I want to round up double to int.
Eg,
double a=0.4, b=0.5;
I want to change them both to integer.
so that
Math.Round
Rounds a double-precision floating-point value to the nearest integral value.
Check out Math.Round. You can then cast the result to an int
.
Use a function in place of MidpointRounding.AwayFromZero
:
myRound(1.11125,4)
Answer:- 1.1114
public static Double myRound(Double Value, int places = 1000)
{
Double myvalue = (Double)Value;
if (places == 1000)
{
if (myvalue - (int)myvalue == 0.5)
{
myvalue = myvalue + 0.1;
return (Double)Math.Round(myvalue);
}
return (Double)Math.Round(myvalue);
places = myvalue.ToString().Substring(myvalue.ToString().IndexOf(".") + 1).Length - 1;
} if ((myvalue * Math.Pow(10, places)) - (int)(myvalue * Math.Pow(10, places)) > 0.49)
{
myvalue = (myvalue * Math.Pow(10, places + 1)) + 1;
myvalue = (myvalue / Math.Pow(10, places + 1));
}
return (Double)Math.Round(myvalue, places);
}
Another option:
string strVal = "32.11"; // will return 33
// string strVal = "32.00" // returns 32
// string strVal = "32.98" // returns 33
string[] valStr = strVal.Split('.');
int32 leftSide = Convert.ToInt32(valStr[0]);
int32 rightSide = Convert.ToInt32(valStr[1]);
if (rightSide > 0)
leftSide = leftSide + 1;
return (leftSide);
Math.Round(0.5) returns zero due to floating point rounding errors, so you'll need to add a rounding error amount to the original value to ensure it doesn't round down, eg.
Console.WriteLine(Math.Round(0.5, 0).ToString()); // outputs 0 (!!)
Console.WriteLine(Math.Round(1.5, 0).ToString()); // outputs 2
Console.WriteLine(Math.Round(0.5 + 0.00000001, 0).ToString()); // outputs 1
Console.WriteLine(Math.Round(1.5 + 0.00000001, 0).ToString()); // outputs 2
Console.ReadKey();
It is simple. So follow this code.
decimal d = 10.5;
int roundNumber = (int)Math.Floor(d + 0.5);
Result is 11