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 result.
For example my double may look something like 77.987654321
, and the two strings conversions convert to to 77.98765
. I need to keep the precision of the value as is.
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.
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.
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
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?
:-)
No guarantees, but try ToString("R").