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Re: Math conundrum

On Dec 20, 2004, at 4:44 AM, Mike Hall wrote:

if it's already a double, then it's already base-2 "imprecise", and
susceptible as such.  Converting it to a String buys you nothing,
though you might do new BigDecimal("1.8");

This was my guess and one other person confirmed off-list. Assuming I have
been forced to a interim calculation double which has resulted in a
"imprecise" base-2 value. Then doing something like String isprecise =
Double.toString(interim); BigDecimal backtoprecise = new
BigDecimal(isprecise); will not work. Because isprecise won't be.

If N is the number which can't be store precisely as a double but can be stored precisely as a decimal, then doing this
double d = N;
BigDecimal b = new BigDecimal(Double.toString(d));
... will only work if Double.toString(d) gets lucky and happens to print the imprecise number back it its precise decimal form. Don't count on it.


An easy but inefficient way to get a numerical (non-string) value of
the form a.b into BigDecimal precisely is to say:    new
BigDecimal(ab).movePointLeft(1)
For example, new BigDecimal(18).movePointLeft(1) is 1.8 exactly.

Is what you are saying that you can sidestep the issue by trying to work
integer or fixed-point somehow and only when done convert back to the
floating-point representation?

A finite-length decimal number is just an integer with a decimal point somewhere. So if your number N *can* be stored precisely as a decimal, then the above will do it in BigDecimal. If N is irrational or has a finite-length longer than 2^32, then you're out of luck in any case, of course.

But you have to consider what you need that precision for: BigDecimal is primarily for explicit math functions that are highly specialized in need. If double's precision is not enough for you, you should very seriously reconsider the approach to your application before going to something as slow as BigDecimal.

Sean

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 >Re: Math conundrum (From: Sean Luke <email@hidden>)
 >Re: Math conundrum (From: "Mike Hall" <email@hidden>)



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