On Mon, 15 Sep 2025 16:50:03 GMT, Joe Darcy <da...@openjdk.org> wrote:
>> fabioromano1 has updated the pull request incrementally with one additional >> commit since the last revision: >> >> Optimization > > HI @fabioromano1, > Ah, I haven't worked on that code for exact divide in a while! > For some historical context, the method was added as part of JSR 13, Decimal > Arithmetic Enhancement, in JDK 5. A fuller derivation of the old bound is > discussed in: > > Fixed, Floating, and Exact Computation with Java's BigDecimalFixed, Floating, > and Exact Computation with Java's BigDecimal > Dr. Dobb's Journal ยท Jun 1, 2004 > https://web.archive.org/web/20210505132021/http://www.drdobbs.com/jvm/fixed-floating-and-exact-computation-wit/184405721 > > At the time I wrote the code, I looked around at the usual sources, Knuth, > etc. and asked around for a bound of decimal digits of 1/b if 1/b is > representable, but didn't find anything. > > I see you're still revising the implementation. Once this settles down, a > review comment from me will be "restore some textual discussion of what the > algorithm is doing." > > For testing, I think it would be good to probe at some values where the old > and new digit estimation techniques differ. > > Also, I think both @rgiulietti and myself look at this PR before it goes > back; I'll adjust the reviewer count accordingly. @jddarcy The technique used here is the following: take $a/b$, compute $b' = b/(2^i 5^j)$, where $i = \max(n \in N \mid b \equiv 0 \mod 2^n)$ and $j = \max(n \in N \mid b \equiv 0 \mod 5^n)$. If $a \not\equiv 0 \mod b'$, then $a/b$ is not a finite decimal number. Otherwise: - if $i \le j$, then $a/b = (a/b')*2^{j-i} * 10^{-j}$; - if $i > j$, then $a/b = (a/b')*5^{j-i} * 10^{-i}$. ------------- PR Comment: https://git.openjdk.org/jdk/pull/27271#issuecomment-3293326363
