C# 将 20 位精度 double 转换为字符串并再次返回

Question

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??

Answer 1

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.

Answer 2

No. Besides the fact that double is binary and therefore not good for decimal values, doubles have a maximum of 15/16 decimal digits (53 bits). decimal has a maximum of 28/29 digits (96 bit), so it would be ok to use it. If you have higher precisions in the database, you need an own bignum class for C#, look in stackoverflow for implementations.