On Tue, 29 Jul 2025 12:06:37 GMT, fabioromano1 <d...@openjdk.org> wrote:
>> 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? > 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. > ``` And this should follow by the fact that `bl - (sh-ex) ≥ bl - sh = bl - (bl - P) = P`, since `ex ≥ 0`. ------------- PR Review Comment: https://git.openjdk.org/jdk/pull/24898#discussion_r2239646258
