Approved! There was a lot of benchmarking work, so I'd like to see the jmh benchmark checked in. Especially since the exact variant we eventually settled on is not in Hacker's Delight.
On Mon, Aug 13, 2018 at 12:09 PM, Ivan Gerasimov <ivan.gerasi...@oracle.com> wrote: > Okay, your last variant is the winner :) > > Here's the updated webrev, benchmark and the graphs: > > http://cr.openjdk.java.net/~igerasim/8209171/02/webrev/ > http://cr.openjdk.java.net/~igerasim/8209171/02/bench/int/MyBenchmark.java > http://cr.openjdk.java.net/~igerasim/8209171/02/bench/int/be > nch-int-04-client.png > http://cr.openjdk.java.net/~igerasim/8209171/02/bench/int/be > nch-int-04-server.png > > With kind regards, > Ivan > > > On 8/13/18 10:10 AM, Ivan Gerasimov wrote: > >> Hi Martin! >> >> Good point about the command line flags, thanks! >> >> These variants are close to numberOfTrailingZeros_07 that I've already >> tested, though you did better by saving one arithmetic operation at the >> return line! >> >> I'll rerun the benchmarks. >> >> With kind regards, >> >> Ivan >> >> >> On 8/13/18 7:56 AM, Martin Buchholz wrote: >> >>> The number of plausible variants is astonishing! >>> >>> --- >>> >>> Your use of -client and -server is outdated, which explains why you get >>> the same results for both (-client is ignored). >>> >>> I'm not sure what's blessed by hotspot team, but for C1 I use >>> -XX:+TieredCompilation -XX:TieredStopAtLevel=1 and for C2 I use >>> -XX:-TieredCompilation -server >>> >>> --- >>> >>> Now I understand the advantage of using ~i & (i - 1): the subsequent >>> zero check is a short-circuit for all odd numbers, better than i & -i, >>> which explains your results - they depend on being able to short-circuit. >>> >>> So just use a more faithful inlining of nlz without trying to improve on >>> it. >>> >>> static int ntz_inlineNlz5(int i) { >>> i = ~i & (i - 1); >>> if (i <= 0) >>> return (i == 0) ? 0 : 32; >>> int n = 1; >>> if (i >= 1 << 16) { n += 16; i >>>= 16; } >>> if (i >= 1 << 8) { n += 8; i >>>= 8; } >>> if (i >= 1 << 4) { n += 4; i >>>= 4; } >>> if (i >= 1 << 2) { n += 2; i >>>= 2; } >>> return n + (i >>> 1); >>> } >>> >>> But it's hard to resist the urge to optimize out a branch: >>> >>> static int ntz_inlineNlz6(int i) { >>> i = ~i & (i - 1); >>> if (i <= 0) return i & 32; >>> int n = 1; >>> if (i >= 1 << 16) { n += 16; i >>>= 16; } >>> if (i >= 1 << 8) { n += 8; i >>>= 8; } >>> if (i >= 1 << 4) { n += 4; i >>>= 4; } >>> if (i >= 1 << 2) { n += 2; i >>>= 2; } >>> return n + (i >>> 1); >>> } >>> >>> >> > -- > With kind regards, > Ivan Gerasimov > >
