# JDK 7u-dev review request 8024356: Double.parseDouble() is slow for long Strings

David Chase david.r.chase at oracle.com
Thu Sep 12 16:32:17 UTC 2013

```I think the reason you use 1075 digits is because it takes 1075 decimal digits behind the decimal point to exactly represent 2^-1075,
and that is 1/2 ulp.  When you discard the tail, if it happens to be all zeroes, I hope you replace it with a zero, not a one, because otherwise you can round incorrectly if the answer is otherwise equal to EVEN + 1/2 ULP.  EVEN + 1/2 ULP rounds to EVEN, EVEN + 1/2 ULP + "1" rounds to EVEN + 1.

Whoops, no, you don't do that.  But I'd call that another bug, and not one worth getting terribly excited about.  Strictly speaking, if your input is decimal_rep_of_half_ulp + a billion zeroes + "1", then the answer is 1 ulp (2^-1074), if the trailing one is missing, then the answer is zero (which is even in the mantissa).

David

On 2013-09-12, at 10:37 AM, David Chase <david.r.chase at oracle.com> wrote:
> On 2013-09-12, at 8:18 AM, Dmitry Nadezhin <dmitry.nadezhin at gmail.com> wrote:
>
>> The reason while binary and decimal digits are mixed can be ullustrated by
>> such example.
>> It is necessary 3 decimal fraction digits to represent exactly pbinary
>> power pow(2,-3) :
>> pow(2,-3)=1/8=0.125
>>
>>
>> On Thu, Sep 12, 2013 at 3:57 PM, David Chase <david.r.chase at oracle.com>wrote:
>>
>>> This explanation seems to combine numbers of binary digits (1075)
>>> and numbers of decimal digits (17), and therefore makes me a little
>>> nervous, though I think 1100 is a conservative choice that, even if not
>>> 100% correct, will be 99.(over 700 9s)% correct.
>>>
>>> David
>
> Yes, but you are adding them.  It is as if you took two lengths, 1075 feet and 17 yards, added them, rounded up, and said that 1100 yards will be adequate (which is entirely true, though perhaps overconservative).  Overconservative is okay.
>
> 2^-1075 is the binary ulp, give or take a fencepost, not the decimal ulp.
>
> David
>

```

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