Finding the number of places after the decimal point of a Double

Question

I have a Double value:

double a = 4.5565;

What is the easiest way to calculate the number of digits after the decimal point (4 in this case

Answer 1

Write a function

int CountDigitsAfterDecimal(double value)
        {
            bool start = false;
            int count = 0;
            foreach (var s in value.ToString())
            {
                if (s == '.')
                {
                    start = true;
                }
                else if (start)
                {
                    count++;
                }
            }

            return count;
        }

Answer 2

base on james answer bat much clearer:

int num = dValue.ToString().Length - (((int)dValue).ToString().Length + 1);

num is the exact number of digits after the decimal point. without including 0 like this(25.520000) in this case, you will get num= 2

Answer 3

I Think String solution is best : ((a-(int)a)+"").length-2

Answer 4

Another solution would be to use some string functions:

private int GetSignificantDecimalPlaces(decimal number, bool trimTrailingZeros = true)
{
  string stemp = Convert.ToString(number);

  if (trimTrailingZeros)
    stemp = stemp.TrimEnd('0');

  return stemp.Length - 1 - stemp.IndexOf(
     Application.CurrentCulture.NumberFormat.NumberDecimalSeparator);
}

Remember to use System.Windows.Forms to get access to Application.CurrentCulture

Answer 5

There's no easy way, especially since the number of digits mathematically speaking might be far more than displayed. For example, 4.5565 is actually stored as 4.556499999999999772626324556767940521240234375 (thanks to harold for calculating that). You're very unlikely to find a useful solution to this problem.

EDIT

You could come up with some algorithm that works like this: if, as you calculate the decimal representation, you find a certain number of 9s (or zeros) in succession, you round up (or down) to the last place before the series of 9s (or zeros) began. I suspect that you would find more trouble down that road than you would anticipate.

Answer 6

I'll perhaps use this code if I needed,

myDoubleNumber.ToString("R").Split('.')[1].Length

"R" here is Round Trip Format Specifier

We need to check for the index bounds first of course.