This is tricky. Some numbers, such as LabVIEW's representation of pi (or if you compute, say, the square root of 2) have an infinite decimal part (don't worry -- LabVIEW only approximates the infinite decimal). So the first question to answer is "How many significant digits to the right of the decimal place do you want?". Suppose this number is N. To take any Double and transform it to have at most N digits to the right of the decimal point, (a) multiply by 10^N, (b) round, (c) divide by 10^N.
Now you can figure out what precision you need. First, get all the significant digits to the right of the decimal point. You can (almost) do this in a While loop with N in a Shift register -- multiply the number by 10 (shifting a decimal from the right to the left of the decimal point), and compare the (original) number with its rounded version, stopping the loop if this is true. Note that this will shift N one too many places (but saves a Case Statement). Now you need to "count the digits" (and subtract 1). Again, While is your Friend -- start dividing by 10, rounding to discard the remainder, and stopping when you get to 0. The index of this second While Loop, when it exits, will be the Precision you want (again, almost).
There are two "gotchas". One is handling negative numbers -- the truncation step can "mis-fire" in the second part of the loop, but you can simply do the computation with the absolute value of N. The second is the case N = 0, which will return a Precision of 0 instead of 1. Simply wrap the second While loop in a Case statement, test (once) for =0 and return 1 if true, else do the While computation.
Suggestion -- make this a sub-VI (call it something clever like "Precision") so you can use it over and over.
BS
