Thanks Chance,
I have read the spec again. From my understanding, the spec requires the
value "x" of finite decimal represented by the Float.toString() meet below
conditions:
1. the "f" is one of float values which are nearest to "x".
2. the number of the fractional part of "x" must be as many as need and the
more digits are needed to uniquely distinguish the argument value.
In our case, if 6.021427E18 is already meet the above conditions and if the
last digit "1" of 6.0214271E18 has no use to distinguish the argument value,
then our result is right and RI is not correct.
In Chance's reply, he has used one example to show this situation. If the
algorithm of our code just do the logic of what spec requires. I believe
harmony does the right things and RI is not correct.
2010/9/16 Shi Jun Zhang <chancedream@gmail.com>
> Hi,
>
> I'm curious whether the implementation in Harmony is wrong or RI just
> doesn't follow the spec.
>
> Here is the spec for java.lang.Float.toString(float f)
>
> public static String *toString*(float f)
>
> Returns a string representation of the float argument. All characters
> mentioned below are ASCII characters.
>
>  If the argument is NaN, the result is the string "NaN".
>  Otherwise, the result is a string that represents the sign and
> magnitude (absolute value) of the argument. If the sign is negative, the
> first character of the result is '' ('\u002D'); if the sign is positive,
> no sign character appears in the result. As for the magnitude *m*:
>  If *m* is infinity, it is represented by the characters "Infinity";
> thus, positive infinity produces the result "Infinity" and negative
> infinity produces the result "Infinity".
>  If *m* is zero, it is represented by the characters "0.0"; thus,
> negative zero produces the result "0.0" and positive zero produces
> the result "0.0".
>  If *m* is greater than or equal to 103 but less than 107, then it
> is represented as the integer part of *m*, in decimal form with no
> leading zeroes, followed by '.' ('\u002E'), followed by one or more
> decimal digits representing the fractional part of *m*.
>  If *m* is less than 103 or greater than or equal to 107, then it is
> represented in socalled "computerized scientific notation." Let
> *n*be the unique integer such that 10
> *n* <= *m* < 10*n*+1; then let *a* be the mathematically exact
> quotient of *m* and 10*n* so that 1 <= *a* < 10. The magnitude is then
> represented as the integer part of *a*, as a single decimal digit,
> followed by '.' ('\u002E'), followed by decimal digits representing
> the fractional part of *a*, followed by the letter 'E' ('\u0045'),
> followed by a representation of *n* as a decimal integer, as produced
> by the method Integer.toString(int).
>
> 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 float. That is,
> suppose that *x* is the exact mathematical value represented by the decimal
> representation produced by this method for a finite nonzero argument *f*.
> Then *f* must be the float value nearest to *x*; or, if two float values
> are
> equally close to *x*, then *f* must be one of them and the least
> significant
> bit of the significand of *f* must be 0.
>
> To create localized string representations of a floatingpoint value, use
> subclasses of NumberFormat.
>
> *Parameters:*f  the float to be converted. *Returns:*a string
> representation of the argument. Please notice the words in red, "as many,
> but only as many". In Mohan's case, the binary value for float
> 6.0214271E18f
> is 01011110 10100111 00100000 11010000 and its exact decimal value is
> 6.021427051003641856E18, should the string representation be "6.021427E18"
> or "6.0214271E18" ?
>
> Let's look at the following 3 closest float value.
> Hex
> Binary Decimal
> 0x5EA720CF 01011110 10100111 00100000 11001111
> 6.021426501247827968E18
> 0x5EA720D0 01011110 10100111 00100000 11010000
> 6.021427051003641856E18
> 0x5EA720D1 01011110 10100111 00100000 11010001
> 6.021427600759455744E18
>
> The argument f is 6.021427051003641856E18. When x is 6.0214271E18, f is the
> float value nearest to x. When x is 6.021427E18, f is still the float value
> nearest to x. When x is 6.02143E18, f is NOT the float value nearest to x.
> So as my understanding, "6.021427E18" is the string should be returned, it
> contains "as many, but only as many" digits to represent the float value f.
> Is there any thing that i misunderstand?
>
> On Wed, Jul 28, 2010 at 1:33 AM, Mohanraj Loganathan
> <mohanraj.l@gmail.com>wrote:
>
> > Following test[1] fails in Harmony. But passes in RI.
> >
> > [1] test:
> > float r=6.0214271E18f;
> > String s=Float.toString(r);
> > assertEquals("6.0214271E18",s);
> > Harmony prints: 6.021427E18 (note '1' is missing from the converted
> > string)
> >
> > When i looked into the implementation of harmony, i come to know its a
> > limitation to the algorithm implemented in Harmony (Printing
> > FloatingPoint Numbers Quickly and Accurately, Robert G. Burger, and
> > R. Kent Dybvig, Programming Language Design and Implementation (PLDI)
> > 1996, pp.108116)
> >
> >
> >
> https://svn.apache.org/repos/asf/harmony/enhanced/java/branches/java6/classlib/modules/luni/src/main/native/luni/shared/dblparse.c
> >
> > function:
> >
> java_org_apache_harmony_luni_util_NumberConverter_bigIntDigitGeneratorInstImpl
> >
> > I am clueless to find the flaw in the implementation. Any comments
> > from experts of this area?
> >
> > I will raise a JIRA for this issue.
> >
> > Thanks and Regards,
> > Mohan
> >
>
>
>
> 
> Shi Jun Zhang
>
