Question

Does the following statement holds for any double (Java primitive double precision IEEE-754) except NaN:

Double.parseDouble(String.valueOf(d)) == d

Said otherwise, does parsing a serialized (using String.valueOf()) double value always yields the exact original double?

Answer 1

With the exception of NaN as you’ve said, yes, that invariant should hold. If not, that’s a JDK bug right there.

Double.toString says this in its Javadoc:

How many digits must be printed for the fractional part of m or a? There must be at least one digit to represent the fractional part, and beyond that as many, but only as many, more digits as are needed to uniquely distinguish the argument value from adjacent values of type double. That is, suppose that x is the exact mathematical value represented by the decimal representation produced by this method for a finite nonzero argument d. Then d must be the double value nearest to x; or if two double values are equally close to x, then d must be one of them and the least significant bit of the significand of d must be 0.

To summarize, it returns enough digits to identify this double uniquely, so Double.parseDouble should return the exact same double that was converted to a string.