C# Converting 20 digit precision double to string and back again

c#, double, precision, tostring

Solution

Use the "R" numeric format string:

double d = 0.00034101243963859839;
string s = d.ToString("R");
//...
double d2 = double.Parse(s);
if(d == d2)
{
  //-- Success
}

The `R` stands for "round-trip". From the linked document:

This format is supported only for the Single and Double types. The round-trip specifier guarantees that a numeric value converted to a string will be parsed back into the same numeric value.

As an aside, I suspect there is no way to keep those last two digits. There's only so much precision available, and I doubt they ever make it into `d` in the first place. But you can make sure your string at least reads back what you do have correctly.

If you really need the additional precision, you might try using a `decimal` instead.

Problem

In C#. I have a double (which I've extracted from a database) that has 20 digit precision. In Visual Studio (using QuickWatch) I can see the value of the double to be = 0.00034101243963859839. I want to display this value in a textbox and then have it be the same value when I take it out and convert it back into a double. But I always lose the last two digits I've tried the following: ``` double d = 0.00034101243963859839; string s = d.ToString(); string s2 = d.ToString("F20"); string s3 = d.ToString("0.00000000000000000000"); -- 20 0's string s4 = (d*100d).ToString(); ``` In these cases: ``` s = 0.000341012439638598 s2 = 0.00034101243963859800 s3 = 0.00034101243963859800 s4 = 0.0341012439638598 ``` I want to be able to do the following: ``` double d = 0.00034101243963859839; string s = d.ToString(); //... double d2 = double.Parse(s); if(d == d2) { //-- Success } ``` Is there any way to keep those last two digits of precision??

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