On Tue, 29 Jul 2025 11:14:15 GMT, Raffaello Giulietti <rgiulie...@openjdk.org> wrote:
>> The proof might help replacing the `if` condition bl - (sh - ex) ≤ ME with >> ex ≤ ME - P. > > To the above, we can also add > > * > * Noting that x ≥ 2^(bl-1) and ex ≥ 0, similarly to the > above we further get > * x 2^(-sh') ≥ 2^(ex+P-1) ≥ 2^(P-1) > * which shows that ⌊x / 2^sh'⌋ has at least P bits of > precision. @rgiulietti Yes, but why that complication, it is not more natural to prove it in the following way? recalling x < 2^bl: x / 2^sh' = x / 2^(sh-ex) ≤ 2^bl / 2^(sh-ex) = 2^(bl - (sh-ex)) ≤ 2^ME < Double.MAX_VALUE ``` relying on the fact that `bl - (sh-ex)` is the length of the shifted value? ------------- PR Review Comment: https://git.openjdk.org/jdk/pull/24898#discussion_r2239589611
