How to format a double with more digits?

Question

I have a double value and I want to convert it to a string with more than the default 15 digits. How can I accomplish this?

(1.23456789987654321d).ToString(); // 1.23456789987654
(12.3456789987654321d).ToString(); // 12.3456789987654
(1.23456789987654321d).ToString("0.######################################"); // 1.23456789987654
(1.23456789987654321d).ToString("0.0000000000000000000000000000000"); // 1.2345678998765400000000000000000

Answer 1

I have a double value and I want to convert it to a string with more than the default 15 digits.

Why? It's basically going to be garbage after 15 digits. You can get the exact value using my DoubleConverter class:

string exact = DoubleConverter.ToExactString(value);

... but after 15 digits, the rest is just noise.

If you want more than 15 significant digits of meaningful data, use decimal .

Answer 2

It is not possible with double as it does not support a precision of more than 15 digits. You can try using decimal data type:

using System;

namespace Code.Without.IDE
{
    public class FloatingTypes
    {
        public static void Main(string[] args)
        {
            decimal deci = 1.23456789987654321M;
            decimal decix = 1.23456789987654321987654321987654321M;
            double doub = 1.23456789987654321d;
            Console.WriteLine(deci); // prints - 1.23456789987654321
            Console.WriteLine(decix); // prints - 1.2345678998765432198765432199
            Console.WriteLine(doub); // prints - 1.23456789987654
        }
    }
}