On Fri, 11 Oct 2024 17:29:59 GMT, Raffaello Giulietti <rgiulie...@openjdk.org> wrote:
>> I'll have to check... >> ... tomorrow ;-) > > IIUC, the code assumes that the floating-point computation > `Math.ceil(intVal.bitLength() * LOG_5_OF_2)` has the same value as the > mathematical ⌈ `intVal.bitLength()` × log5(2) ⌉. > I don't think this is safe, as it might happen that the computed and > mathematical values are off by ±1. > To ensure 5^`maxPowsOf5` >= `intVal` (that is, maxPowsOf5 >= log5(intVal)) it > would be more prudent to have > > long maxPowsOf5 = (long) Math.ceil(intVal.bitLength() * LOG_5_OF_2) + 1; > > > But I think what you really want is maybe to meet 5^`maxPowsOf5` <= `intVal` > < 5^(`maxPowsOf5` + 1) instead? Actually, if we reason in terms of "ulp vs precision", the computation should be safe: `Math.log()`'s results are within 1 ulp of the exact result, and the floating point operations are a multiplication and a division. The division to compute `LOG_5_OF_2` costs 1/2 ulp plus the errors of the operands, so 2.5 ulps. Same for multiplication, but `intVal.bitLength()` has an exact `double` value, so the total roundoff error of `intVal.bitLength() * LOG_5_OF_2` is 3 ulps. Since the integer part of `intVal.bitLength() * LOG_5_OF_2` is representable with 31 bits, and double has 53 bits of precision, we can reasonably say that `Math.ceil()` can always guarantee `maxPowsOf5 >= log5(intVal)`. ------------- PR Review Comment: https://git.openjdk.org/jdk/pull/21323#discussion_r1797292843
