了解jvm中的循环性能 | 了解jvm中的循环性能

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问题描述

我正在玩 jmh 以及关于

IF =取指令，ID =指令解码，EX =执行，MEM =存储器访问，WB =寄存器回写

来自：

有关流水线的更多信息：

I'm playing with jmh and in the section about looping they said that

I tried it myself

    @Benchmark
    @OperationsPerInvocation(1)
    public int measurewrong_1() {
        return reps(1);
    }

    @Benchmark
    @OperationsPerInvocation(1000)
    public int measurewrong_1000() {
        return reps(1000);
    }

and got the following result:

Benchmark                      Mode  Cnt  Score    Error  Units
MyBenchmark.measurewrong_1     avgt   15  2.425 ±  0.137  ns/op
MyBenchmark.measurewrong_1000  avgt   15  0.036 ±  0.001  ns/op

It indeed shows that the MyBenchmark.measurewrong_1000 is dramatically faster than MyBenchmark.measurewrong_1. But I cannot really understand the optimization JVM does to make this performance improvement.

What do they mean the loop is unrolled/pipelined?

解决方案

Loop unrolling makes pipelining possible. So the pipeline-able CPU (for example RISC) can execute the unrolled code in parallel.

So if your CPU is able to execute 5 pipelines in parallel, your loop will be unrolled in the way:

// pseudo code
int pipelines = 5;
for(int i = 0; i < length; i += pipelines){
    s += (x + y);
    s += (x + y);
    s += (x + y);
    s += (x + y);
    s += (x + y);
}

IF = Instruction Fetch, ID = Instruction Decode, EX = Execute, MEM = Memory access, WB = Register write back

From Oracle White paper:

more information about pipelining: Classic RISC pipeline

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