If you really have a number with 52 decimal digits of precision
required, you can not store it in a double.
I believe that a double is only good for about 16-17 digits.
What is the origin of such a number and what do you want to do with it?
You will have to write all of your own arithmetic or find a package
designed to work with large numbers with high precision.
There are probably ways to deal with this but ordinary float or double
will not do it even in JVM. You may get one test to work but there is no
assurance that with a different sequence of operations with different
numbers you will not get a different result since the hardware is
throwing away the extra precision.
Ron
elibol wrote:
> Thank you Jim.
>
> To clarify the problem we are discussing, when a string is parsed
into a
> number, and the string is representing a very large number, the number
> does
> not yield the expected results when used in an operation. This is
> demonstrated in the first modulo experiment. This issue does not exist
in
> AVM1 or JVM.
>
> "From previous data packing experiments, I've found that the last
nibble
> (4 bits) of a 64-bit Number isn't reliable when you're casting to and
> from Numbers (e.g . reading 8 bytes out of a ByteArray into a
Number or
> doing round-trips via the toString and parseFloat methods). I'm not
> sure why this is the case--perhaps someone from Adobe can reply and
> speak to this particular issue."
>
> This seems relevant, as 6.3e+51 would require an allocation of 3 64
bit
> Numbers (approx. 173) to be represented, using at least 2 sets of
those
4
> unreliable bits you've mentioned. Is this correct?
>
> Is there a solution/technique for correctly representing such parsed
> Numbers? The subject Number will be concerned with properly
> representing its
> value in order to be used in any operation supported by as3.
>
> Is it impossible to write a parseFloat function that would correctly
> parse
> Numbers under the new Number specification?
>
> Thank you Fumio and Jim.
>
> Note: I posted this 2 days ago and I didn't notice that I got an
> Undeliverable error. I've been getting these a lot. Not sure what the
> deal
> is...
>
> On 3/7/07, elibol <[EMAIL PROTECTED]> wrote:
>>
>> To clarify the problem we are discussing, when a string is parsed
into
a
>> number, and the string is representing a very large number, the
>> number does
>> not yield the expected results when used in an operation. This is
>> demonstrated in the first modulo experiment. This issue does not
>> exist in
>> AVM1 or JVM.
>>
>> On 3/7/07, elibol <[EMAIL PROTECTED]> wrote:
>> >
>> > Thank you Jim.
>> >
>> > "From previous data packing experiments, I've found that the last
>> nibble
>> > (4 bits) of a 64-bit Number isn't reliable when you're casting
to and
>> > from Numbers (e.g . reading 8 bytes out of a ByteArray into a
>> Number or
>> > doing round-trips via the toString and parseFloat methods). I'm
not
>> > sure why this is the case--perhaps someone from Adobe can reply and
>> > speak to this particular issue."
>> >
>> > This seems relevant, as 6.3e+51 would require an allocation of 3 64
>> bit
>> > Numbers (approx. 173) to be represented, using at least 2 sets of
>> those 4
>> > unreliable bits you've mentioned. Is this correct?
>> >
>> > Is there a solution/technique for correctly representing such
parsed
>> > Numbers? The subject Number will be concerned with properly
>> representing its
>> > value in order to be used in any operation supported by as3.
>> >
>> > Is it impossible to write a parseFloat function that would
correctly
>> > parse Numbers under this specification?
>> >
>> > Thank you Fumio and Jim.
>> >
>> > On 3/6/07, Jim Cheng <[EMAIL PROTECTED]> wrote:
>> > >
>> > > Fumio Nonaka wrote:
>> > > > 2 floating point numbers are NOT "close enough". That IS the
>> > > problem.
>> > > >
>> > > > var _str:String = "1.2e+51";
>> > > > var n:Number = parseFloat(_str);
>> > > > trace(( n-1.2e+51 ) >
100000000000000000000000000000000000); //
>> > > true
>> > >
>> > > In ActionScript 3, the native Number data type is internally
>> > > represented
>> > > as a IEEE-754 double-precision floating-point number. Due to the
>> way
>> > > that the IEEE-754 standard defines how the bits are used to
>> represent
>> > > the number, the accuracy of the mantissa precision (always 52
>> bits, or
>> > > 16 decimal digits) changes depending on the exponent (always 11
>> bits).
>> > >
>> > > See: http://en.wikipedia.org/wiki/IEEE_754
>> > >
>> > > This is to say, the "close enough" value that you need to compare
>> the
>> > > absolute difference between the two Numbers scales in magnitude
with
>> > > the
>> > > exponent. This can be particularly bad if you need arbitrary
>> > > precision,
>> > > (e.g. when doing financial or scientific calcuations), as while
>> > > 1.32e+36
>> > > is paltry compared to 1.2e+51, no one would want to be swindled
>> out of
>> > > 1.32e+36 dollars due to faulty floating point comparisons, hence
the
>> > > need for arbitrary precision integer libraries for such
applications
>> > > as
>> > > was recently mentioned on this list.
>> > >
>> > > Fortunately however, if you don't need this kind of exact
precision,
>> > > but
>> > > simply need to match large values originally parsed from
strings to
>> > > Numbers as per your example, there is a much better and easier
>> way to
>> > > compare very large Numbers in ActionScript 3--by inspecting
them at
>> > > the
>> > > bit level following the IEEE-754 specification.
>> > >
>> > > From previous data packing experiments, I've found that the last
>> > > nibble
>> > > (4 bits) of a 64-bit Number isn't reliable when you're casting to
>> and
>> > > from Numbers (e.g. reading 8 bytes out of a ByteArray into a
>> Number or
>> > > doing round-trips via the toString and parseFloat methods). I'm
not
>> > > sure why this is the case--perhaps someone from Adobe can
reply and
>> > > speak to this particular issue.
>> > >
>> > > That being said, all you really need to do for a nearly-equals
>> Number
>> > > comparison is a byte-by-byte comparison save for the last
byte, in
>> > > which
>> > > case you only compare the four most significant bits. If all
bits
>> > > aside
>> > > from the last nibble match, the Numbers are close enough.
>> > >
>> > > Here's how I do it:
>> > >
>> > > <code>
>> > >
>> > > /**
>> > > * Compare two ActionScript 3 Numbers for near-equality.
>> > > *
>> > > * @param The first Number
>> > > * @param The second Number
>> > > *
>> > > * @return True on near-equality, false otherwise.
>> > > */
>> > > public function closeEnough(a:Number, b:Number):Boolean {
>> > > var ba:ByteArray, i:uint;
>> > > if (a == b) {
>> > > // A is explicitly equal to B
>> > > return true;
>> > > }
>> > > else {
>> > > // A isn't exactly equal to B, so we need to
>> > > // check the Numbers' bytes one byte at a time.
>> > > ba = new ByteArray();
>> > > ba.writeDouble(a);
>> > > ba.writeDouble(b);
>> > >
>> > > // If any of the first 7 bytes differ, then
>> > > // the two values are not close enough.
>> > > for (i = 0; i < 7; i++) {
>> > > if (ba[i] != ba[i + 8]) {
>> > > return false;
>> > > }
>> > > }
>> > >
>> > > // Mask the last four bits out and compare the
>> > > // last byte. If they match, the Numbers are
>> > > // close enough. The last nibble tends not to
>> > > // be reliable enough for comparison, so we
>> > > // allow these to differ and the two Numbers
>> > > // still be considered close enough.
>> > > if ((ba[7] & 0xf0) != (ba[15] & 0xf0)) {
>> > > return false;
>> > > }
>> > > }
>> > > return true;
>> > > }
>> > >
>> > > </code>
>> > >
>> > > Jim Cheng
>> > > effectiveUI
>> > > _______________________________________________
>> > > [email protected]
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>> > >
>> >
>> >
>>
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