Why Is ToString() Rounding My Double Value?

Question

How do I prevent my double value from being rounded when converting to a string? I have tried both Convert.ToString and ToString() with the same r

Answer 1

No guarantees, but try ToString("R").

Answer 2

Jon Skeet's DoubleConverter class has a ToExactString() method which will return the exact value of the double as a string.

http://www.yoda.arachsys.com/csharp/DoubleConverter.cs

Answer 3

Try something like

myDouble.ToString("R")

See also The Round-trip ("R") Format Specifier:

The round-trip ("R") format specifier guarantees that a numeric value that is converted to a string will be parsed back into the same numeric value.

Answer 4

I would assume that the main answer for rounding away the last two digits, is to hide numerical instability/rounding due to float/double finite precision.

Example with no rounding:

(Math.Sqrt(7)).ToString("G17") = "2.6457513110645907"

(Math.Sqrt(7)+6).ToString("G17") = "8.6457513110645898"

Looks a bit strange in the last 3 digits, right?

Example with rounding:

(Math.Sqrt(7)).ToString() = "2.64575131106459"

(Math.Sqrt(7)+6).ToString() = "8.64575131106459"

Look "perfect", right?

:-)

Answer 5

By default the .ToString() method of Double returns 15 digits of precision. If you want the full 17 digits that the double value holds internally, you need to pass the "G17" format specifier to the method.

String s = value.ToString("G17");

Sourced from the MSDN docs:

By default, the return value only contains 15 digits of precision although a maximum of 17 digits is maintained internally. If the value of this instance has greater than 15 digits, ToString returns PositiveInfinitySymbol or NegativeInfinitySymbol instead of the expected number. If you require more precision, specify format with the "G17" format specification, which always returns 17 digits of precision, or "R", which returns 15 digits if the number can be represented with that precision or 17 digits if the number can only be represented with maximum precision.