Java 8u40 Math.round() 非常慢

2022-09-02 03:55:55

我有一个用Java 8编写的相当简单的业余爱好项目,它在其操作模式之一中广泛使用重复的Math.round()调用。例如,一种这样的模式通过 ExecutorService 生成 4 个线程,并将 48 个可运行的任务排入队列,每个任务运行类似于以下代码块的内容 2^31 次:

int3 = Math.round(float1 + float2);
int3 = Math.round(float1 * float2);
int3 = Math.round(float1 / float2);

事实并非如此(涉及数组和嵌套循环),但你明白了。无论如何,在Java 8u40之前,类似于上述代码的代码可以在大约13秒内在AMD A10-7700k上完成约1030亿个指令块的完整运行。使用Java 8u40,做同样的事情大约需要260秒。没有对代码进行任何更改,什么都没有,只是Java更新。

有没有人注意到Math.round()变慢了,特别是当它被重复使用时?这几乎就像JVM之前正在做某种优化,它不再做了。也许它在8u40之前使用SIMD,现在不是?

编辑:我已经完成了对MVCE的第二次尝试。您可以在此处下载第一次尝试:

https://www.dropbox.com/s/rm2ftcv8y6ye1bi/MathRoundMVCE.zip?dl=0

第二次尝试如下。我的第一次尝试已经从这篇文章中删除了,因为它被认为太长了,并且容易受到JVM的死代码删除优化(显然在8u40中发生得更少)。

import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;

public class MathRoundMVCE
{           
    static long grandtotal = 0;
    static long sumtotal = 0;

    static float[] float4 = new float[128];
    static float[] float5 = new float[128];
    static int[] int6 = new int[128];
    static int[] int7 = new int[128];
    static int[] int8 = new int[128];
    static long[] longarray = new long[480];

    final static int mil = 1000000;

    public static void main(String[] args)
    {       
        initmainarrays();
        OmniCode omni = new OmniCode();
        grandtotal = omni.runloops() / mil;
        System.out.println("Total sum of operations is " + sumtotal);
        System.out.println("Total execution time is " + grandtotal + " milliseconds");
    }   

    public static long siftarray(long[] larray)
    {
        long topnum = 0;
        long tempnum = 0;
        for (short i = 0; i < larray.length; i++)
        {
            tempnum = larray[i];
            if (tempnum > 0)
            {
                topnum += tempnum;
            }
        }
        topnum = topnum / Runtime.getRuntime().availableProcessors();
        return topnum;
    }

    public static void initmainarrays()
    {
        int k = 0;

        do
        {           
            float4[k] = (float)(Math.random() * 12) + 1f;
            float5[k] = (float)(Math.random() * 12) + 1f;
            int6[k] = 0;

            k++;
        }
        while (k < 128);        
    }       
}

class OmniCode extends Thread
{           
    volatile long totaltime = 0;
    final int standard = 16777216;
    final int warmup = 200000;

    byte threads = 0;

    public long runloops()
    {
        this.setPriority(MIN_PRIORITY);

        threads = (byte)Runtime.getRuntime().availableProcessors();
        ExecutorService executor = Executors.newFixedThreadPool(threads);

        for (short j = 0; j < 48; j++)
        {           
            executor.execute(new RoundFloatToIntAlternate(warmup, (byte)j));
        }

        executor.shutdown();

        while (!executor.isTerminated())
        {
            try
            {
                Thread.sleep(100);
            } 
            catch (InterruptedException e)
            {
                //Do nothing                
            }
        }

        executor = Executors.newFixedThreadPool(threads);

        for (short j = 0; j < 48; j++)
        {           
            executor.execute(new RoundFloatToIntAlternate(standard, (byte)j));          
        }

        executor.shutdown();

        while (!executor.isTerminated())
        {
            try
            {
                Thread.sleep(100);
            } 
            catch (InterruptedException e)
            {
                //Do nothing                
            }
        }

        totaltime = MathRoundMVCE.siftarray(MathRoundMVCE.longarray);   

        executor = null;
        Runtime.getRuntime().gc();
        return totaltime;
    }
}

class RoundFloatToIntAlternate extends Thread
{       
    int i = 0;
    int j = 0;
    int int3 = 0;
    int iterations = 0;
    byte thread = 0;

    public RoundFloatToIntAlternate(int cycles, byte threadnumber)
    {
        iterations = cycles;
        thread = threadnumber;
    }

    public void run()
    {
        this.setPriority(9);
        MathRoundMVCE.longarray[this.thread] = 0;
        mainloop();
        blankloop();    

    }

    public void blankloop()
    {
        j = 0;
        long timer = 0;
        long totaltimer = 0;

        do
        {   
            timer = System.nanoTime();
            i = 0;

            do
            {
                i++;
            }
            while (i < 128);
            totaltimer += System.nanoTime() - timer;            

            j++;
        }
        while (j < iterations);         

        MathRoundMVCE.longarray[this.thread] -= totaltimer;
    }

    public void mainloop()
    {
        j = 0;
        long timer = 0; 
        long totaltimer = 0;
        long localsum = 0;

        int[] int6 = new int[128];
        int[] int7 = new int[128];
        int[] int8 = new int[128];

        do
        {   
            timer = System.nanoTime();
            i = 0;

            do
            {
                int6[i] = Math.round(MathRoundMVCE.float4[i] + MathRoundMVCE.float5[i]);
                int7[i] = Math.round(MathRoundMVCE.float4[i] * MathRoundMVCE.float5[i]);
                int8[i] = Math.round(MathRoundMVCE.float4[i] / MathRoundMVCE.float5[i]);

                i++;
            }
            while (i < 128);
            totaltimer += System.nanoTime() - timer;

            for(short z = 0; z < 128; z++)
            {
                localsum += int6[z] + int7[z] + int8[z];
            }       

            j++;
        }
        while (j < iterations);         

        MathRoundMVCE.longarray[this.thread] += totaltimer;
        MathRoundMVCE.sumtotal = localsum;
    }
}

长话短说,这段代码在8u25中的执行速度与在8u40中的执行速度大致相同。如您所见,我现在将所有计算结果记录到数组中,然后将循环的定时部分之外的这些数组求和为局部变量,然后将其写入外部循环末尾的静态变量。

低于 8u25:总执行时间为 261545 毫秒

低于 8u40:总执行时间为 266890 毫秒

测试条件与以前相同。因此,看起来 8u25 和 8u31 正在执行 8u40 停止执行的死代码删除,从而导致代码在 8u40 中“变慢”。这并不能解释每一件奇怪的小事,但这似乎是其中的大部分。作为额外的奖励,这里提供的建议和答案给了我灵感,以改善我的爱好项目的其他部分,对此我非常感激。谢谢大家!


答案 1

随意的基准测试:你对A进行基准测试,但实际上测量了B,并得出结论你已经测量了C。

现代JVM过于复杂,可以进行各种优化。如果你试图测量一些小段代码,在没有非常非常详细的JVM正在做什么的情况下正确地做到这一点真的很复杂。许多基准测试的罪魁祸首是死代码消除:编译器足够聪明,可以推断出一些计算是多余的,并完全消除它们。请阅读以下幻灯片 http://shipilev.net/talks/jvmls-July2014-benchmarking.pdf。为了“修复”Adam的微基准标记(我仍然无法理解它在测量什么,并且这个“修复”没有考虑到预热,OSR和许多其他微平台标记陷阱),我们必须将计算结果打印到系统输出:

    int result = 0;
    long t0 = System.currentTimeMillis();
    for (int i = 0; i < 1e9; i++) {
        result += Math.round((float) i / (float) (i + 1));
    }
    long t1 = System.currentTimeMillis();
    System.out.println("result = " + result);
    System.out.println(String.format("%s, Math.round(float), %.1f ms", System.getProperty("java.version"), (t1 - t0)/1f));

因此:

result = 999999999
1.8.0_25, Math.round(float), 5251.0 ms

result = 999999999
1.8.0_40, Math.round(float), 3903.0 ms

原始 MVCE 示例的相同“修复”

It took 401772 milliseconds to complete edu.jvm.runtime.RoundFloatToInt. <==== 1.8.0_40

It took 410767 milliseconds to complete edu.jvm.runtime.RoundFloatToInt. <==== 1.8.0_25

如果你想衡量数学#round的实际成本,你应该写这样的东西(基于jmh)

package org.openjdk.jmh.samples;

import org.openjdk.jmh.annotations.*;
import org.openjdk.jmh.runner.Runner;
import org.openjdk.jmh.runner.RunnerException;
import org.openjdk.jmh.runner.options.Options;
import org.openjdk.jmh.runner.options.OptionsBuilder;
import org.openjdk.jmh.runner.options.VerboseMode;

import java.util.Random;
import java.util.concurrent.TimeUnit;

@State(Scope.Benchmark)
@BenchmarkMode(Mode.AverageTime)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
@Warmup(iterations = 3, time = 5, timeUnit = TimeUnit.SECONDS)
@Measurement(iterations = 3, time = 5, timeUnit = TimeUnit.SECONDS)
public class RoundBench {

    float[] floats;
    int i;

    @Setup
    public void initI() {
        Random random = new Random(0xDEAD_BEEF);
        floats = new float[8096];
        for (int i = 0; i < floats.length; i++) {
            floats[i] = random.nextFloat();
        }
    }

    @Benchmark
    public float baseline() {
        i++;
        i = i & 0xFFFFFF00;
        return floats[i];
    }

    @Benchmark
    public int round() {
        i++;
        i = i & 0xFFFFFF00;
        return Math.round(floats[i]);
    }

    public static void main(String[] args) throws RunnerException {
        Options options = new OptionsBuilder()
                .include(RoundBench.class.getName())
                .build();
        new Runner(options).run();
    }
}

我的结果是:

1.8.0_25
Benchmark            Mode  Cnt  Score   Error  Units
RoundBench.baseline  avgt    6  2.565 ± 0.028  ns/op
RoundBench.round     avgt    6  4.459 ± 0.065  ns/op

1.8.0_40 
Benchmark            Mode  Cnt  Score   Error  Units
RoundBench.baseline  avgt    6  2.589 ± 0.045  ns/op
RoundBench.round     avgt    6  4.588 ± 0.182  ns/op

为了找到问题的根本原因,您可以使用 https://github.com/AdoptOpenJDK/jitwatch/。为了节省时间,我可以说Math#round的JITted代码的大小在8.0_40中增加了。对于小方法来说,它几乎是不明显的,但是在大型方法的情况下,太长的机器代码表会污染指令缓存。


答案 2

基于OP的MVCE

  • 可以进一步简化
  • 将语句更改为,以减少删除死代码的几率。 从 8u31 到 8u40 的差异慢 3 倍。使用差异仅慢 15%。int3 =int3 +=int3 =int3 +=
  • 打印结果,进一步降低死码删除优化的机会

法典

public class MathTime {
    static float[][] float1 = new float[8][16];
    static float[][] float2 = new float[8][16];

    public static void main(String[] args) {
        for (int j = 0; j < 8; j++) {
            for (int k = 0; k < 16; k++) {
                float1[j][k] = (float) (j + k);
                float2[j][k] = (float) (j + k);
            }
        }
        new Test().run();
    }

    private static class Test {
        int int3;

        public void run() {
            for (String test : new String[] { "warmup", "real" }) {

                long t0 = System.nanoTime();

                for (int count = 0; count < 1e7; count++) {
                    int i = count % 8;
                    int3 += Math.round(float1[i][0] + float2[i][0]);
                    int3 += Math.round(float1[i][1] + float2[i][1]);
                    int3 += Math.round(float1[i][2] + float2[i][2]);
                    int3 += Math.round(float1[i][3] + float2[i][3]);
                    int3 += Math.round(float1[i][4] + float2[i][4]);
                    int3 += Math.round(float1[i][5] + float2[i][5]);
                    int3 += Math.round(float1[i][6] + float2[i][6]);
                    int3 += Math.round(float1[i][7] + float2[i][7]);
                    int3 += Math.round(float1[i][8] + float2[i][8]);
                    int3 += Math.round(float1[i][9] + float2[i][9]);
                    int3 += Math.round(float1[i][10] + float2[i][10]);
                    int3 += Math.round(float1[i][11] + float2[i][11]);
                    int3 += Math.round(float1[i][12] + float2[i][12]);
                    int3 += Math.round(float1[i][13] + float2[i][13]);
                    int3 += Math.round(float1[i][14] + float2[i][14]);
                    int3 += Math.round(float1[i][15] + float2[i][15]);

                    int3 += Math.round(float1[i][0] * float2[i][0]);
                    int3 += Math.round(float1[i][1] * float2[i][1]);
                    int3 += Math.round(float1[i][2] * float2[i][2]);
                    int3 += Math.round(float1[i][3] * float2[i][3]);
                    int3 += Math.round(float1[i][4] * float2[i][4]);
                    int3 += Math.round(float1[i][5] * float2[i][5]);
                    int3 += Math.round(float1[i][6] * float2[i][6]);
                    int3 += Math.round(float1[i][7] * float2[i][7]);
                    int3 += Math.round(float1[i][8] * float2[i][8]);
                    int3 += Math.round(float1[i][9] * float2[i][9]);
                    int3 += Math.round(float1[i][10] * float2[i][10]);
                    int3 += Math.round(float1[i][11] * float2[i][11]);
                    int3 += Math.round(float1[i][12] * float2[i][12]);
                    int3 += Math.round(float1[i][13] * float2[i][13]);
                    int3 += Math.round(float1[i][14] * float2[i][14]);
                    int3 += Math.round(float1[i][15] * float2[i][15]);

                    int3 += Math.round(float1[i][0] / float2[i][0]);
                    int3 += Math.round(float1[i][1] / float2[i][1]);
                    int3 += Math.round(float1[i][2] / float2[i][2]);
                    int3 += Math.round(float1[i][3] / float2[i][3]);
                    int3 += Math.round(float1[i][4] / float2[i][4]);
                    int3 += Math.round(float1[i][5] / float2[i][5]);
                    int3 += Math.round(float1[i][6] / float2[i][6]);
                    int3 += Math.round(float1[i][7] / float2[i][7]);
                    int3 += Math.round(float1[i][8] / float2[i][8]);
                    int3 += Math.round(float1[i][9] / float2[i][9]);
                    int3 += Math.round(float1[i][10] / float2[i][10]);
                    int3 += Math.round(float1[i][11] / float2[i][11]);
                    int3 += Math.round(float1[i][12] / float2[i][12]);
                    int3 += Math.round(float1[i][13] / float2[i][13]);
                    int3 += Math.round(float1[i][14] / float2[i][14]);
                    int3 += Math.round(float1[i][15] / float2[i][15]);

                }
                long t1 = System.nanoTime();
                System.out.println(int3);
                System.out.println(String.format("%s, Math.round(float), %s, %.1f ms", System.getProperty("java.version"), test, (t1 - t0) / 1e6));
            }
        }
    }
}

结果

adam@brimstone:~$ ./jdk1.8.0_40/bin/javac MathTime.java;./jdk1.8.0_40/bin/java -cp . MathTime 
1.8.0_40, Math.round(float), warmup, 6846.4 ms
1.8.0_40, Math.round(float), real, 6058.6 ms
adam@brimstone:~$ ./jdk1.8.0_31/bin/javac MathTime.java;./jdk1.8.0_31/bin/java -cp . MathTime 
1.8.0_31, Math.round(float), warmup, 5717.9 ms
1.8.0_31, Math.round(float), real, 5282.7 ms
adam@brimstone:~$ ./jdk1.8.0_25/bin/javac MathTime.java;./jdk1.8.0_25/bin/java -cp . MathTime 
1.8.0_25, Math.round(float), warmup, 5702.4 ms
1.8.0_25, Math.round(float), real, 5262.2 ms

观察

  • 对于Math.round(float)的琐碎使用,我在我的平台(Linux x86_64)上找不到性能差异。基准测试只有差异,我之前的天真和不正确的基准测试只暴露了Ivan的答案和Marco13的评论所指出的那样,优化行为的差异。
  • 8u40在死代码消除方面比以前的版本更不激进,这意味着在某些极端情况下执行更多的代码,因此速度较慢。
  • 8u40需要稍长的时间才能预热,但一旦“在那里”,速度会更快。

来源分析

令人惊讶的是,Math.round(float)是一个纯Java实现而不是本机,8u31和8u40的代码是相同的。

diff  jdk1.8.0_31/src/java/lang/Math.java jdk1.8.0_40/src/java/lang/Math.java
-no differences-

public static int round(float a) {
    int intBits = Float.floatToRawIntBits(a);
    int biasedExp = (intBits & FloatConsts.EXP_BIT_MASK)
            >> (FloatConsts.SIGNIFICAND_WIDTH - 1);
    int shift = (FloatConsts.SIGNIFICAND_WIDTH - 2
            + FloatConsts.EXP_BIAS) - biasedExp;
    if ((shift & -32) == 0) { // shift >= 0 && shift < 32
        // a is a finite number such that pow(2,-32) <= ulp(a) < 1
        int r = ((intBits & FloatConsts.SIGNIF_BIT_MASK)
                | (FloatConsts.SIGNIF_BIT_MASK + 1));
        if (intBits < 0) {
            r = -r;
        }
        // In the comments below each Java expression evaluates to the value
        // the corresponding mathematical expression:
        // (r) evaluates to a / ulp(a)
        // (r >> shift) evaluates to floor(a * 2)
        // ((r >> shift) + 1) evaluates to floor((a + 1/2) * 2)
        // (((r >> shift) + 1) >> 1) evaluates to floor(a + 1/2)
        return ((r >> shift) + 1) >> 1;
    } else {
        // a is either
        // - a finite number with abs(a) < exp(2,FloatConsts.SIGNIFICAND_WIDTH-32) < 1/2
        // - a finite number with ulp(a) >= 1 and hence a is a mathematical integer
        // - an infinity or NaN
        return (int) a;
    }
}

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