All began with these simple lines of code:
string s = "16.9";
double d = Convert.ToDouble(s);
d*=100;
The result should be 1690.0, but it's not. d is equal to 1689.9999999999998. All I want to do is to round a double to value with 2 digit after decimal separator. Here is my function.
private double RoundFloat(double Value)
{
float sign = (Value < 0) ? -0.01f : 0.01f;
if (Math.Abs(Value) < 0.00001) Value = 0;
string SVal = Value.ToString();
string DecimalSeparator = System.Globalization.CultureInfo.CurrentCulture.NumberFormat.CurrencyDecimalSeparator;
int i = SVal.IndexOf(DecimalSeparator);
if (i > 0)
{
int SRnd;
try
{
// вземи втората цифра след десетичния разделител
SRnd = Convert.ToInt32(SVal.Substring(i + 3, 1));
}
catch
{
SRnd = 0;
}
if (SVal.Length > i + 3)
SVal = SVal.Substring(0, i + 3);
//SVal += "00001";
try
{
double result = (SRnd >= 5) ? Convert.ToDouble(SVal) + sign : Convert.ToDouble(SVal);
//result = Math.Round(result, 2);
return result;
}
catch
{
return 0;
}
}
else
{
return Value;
}
But again the same problem, converting from string to double is not working as I want. A workaround to this problem is to concatenate "00001" to the string and then use the Math.Round function (commented in the example above).
This double value multiplied to 100 (as integer) is send to a device (cash register) and this values must be correct.
I am using VS2005 + .NET CF 2.0
Is there another more "elegant" solution, I am not happy with this one.
Doubles can't exactly represent 16.9. I suggest you convert it to
decimal
instead:You might just want to keep using the
decimal
instead of thedouble
, since you say you'll be using it for monetary purposes. Remember thatdecimal
is intended to exactly represent decimal numbers that fit in its precision, whiledouble
will only exactly represent binary numbers that do.The best solution for not going be crazy is: string s = "16.9";
For ,/. double d = Convert.ToDouble(s.Replace(',','.'),System.Globalization.CultureInfo.InvariantCulture);
For rounding: Convert.ToDouble((d).ToString("F2"));
Edit I got the wrong question - the rounding problems are inherent to a floating point type (float, double). You should use decimal for this.