根据其文档，System.nanoTime 返回 自某个固定但任意的原始时间以来的纳秒数。但是，在我尝试了以下代码的所有 x64 机器上，都存在时间跳跃，移动了固定 原始时间。我使用替代方法(此处为 currentTimeMillis)获取正确时间的方法可能存在一些缺陷。然而，测量相对时间(持续时间)的主要目的也受到负面影响。

在将不同的队列与 LMAX 的 Disruptor 进行比较时，我遇到了这个问题，试图测量延迟，有时我会遇到非常负的延迟。在这些情况下，开始和结束时间戳由不同的线程创建，但延迟是在这些线程完成后计算的。

我这里的代码使用 nanoTime 来计算时间，计算 currentTimeMillis 时间的固定原点，并比较调用之间的原点。因为我必须在这里问一个问题:这段代码有什么问题？为什么它观察到违反固定来源契约(Contract)的行为？还是不是？

import java.text.*;

/**
 * test coherency between {@link System#currentTimeMillis()} and {@link System#nanoTime()}
 */
public class TimeCoherencyTest {

    static final int MAX_THREADS = Math.max( 1, Runtime.getRuntime().availableProcessors() - 1);
    static final long RUNTIME_NS = 1000000000L * 100;
    static final long BIG_OFFSET_MS = 2;

    static long startNanos;
    static long firstNanoOrigin;
    static {
        initNanos();
    }

    private static void initNanos() {
        long    millisBefore = System.currentTimeMillis();
        long    millisAfter;
        do {
            startNanos = System.nanoTime();
            millisAfter = System.currentTimeMillis();
        } while ( millisAfter != millisBefore);
        firstNanoOrigin = ( long) ( millisAfter - ( startNanos / 1e6));
    }

    static NumberFormat lnf = DecimalFormat.getNumberInstance();
    static {
        lnf.setMaximumFractionDigits( 3);
        lnf.setGroupingUsed( true);
    };

    static class TimeCoherency {
        long    firstOrigin;
        long    lastOrigin;
        long    numMismatchToLast = 0;
        long    numMismatchToFirst = 0;
        long    numMismatchToFirstBig = 0;
        long    numChecks = 0;

        public TimeCoherency( long firstNanoOrigin) {
            firstOrigin = firstNanoOrigin;
            lastOrigin = firstOrigin;
        }
    }

    public static void main( String[] args) {
        Thread[]    threads = new Thread[ MAX_THREADS];
        for ( int i = 0;  i < MAX_THREADS;  i++) {
            final int   fi = i;
            final TimeCoherency tc = new TimeCoherency( firstNanoOrigin);
            threads[ i] = new Thread() {
                @Override
                public void run() {
                    long    start = getNow( tc);
                    long    firstOrigin = tc.lastOrigin;    // get the first origin for this thread
                    System.out.println( "Thread " + fi + " started at " + lnf.format( start) + " ns");
                    long    nruns = 0;
                    while ( getNow( tc) < RUNTIME_NS) {
                        nruns++;
                    }
                    final long  runTimeNS = getNow( tc) - start;
                    final long  originDrift = tc.lastOrigin - firstOrigin;
                    nruns += 3; // account for start and end call and the one that ends the loop
                    final long skipped = nruns - tc.numChecks;
                    System.out.println( "Thread " + fi + " finished after " + lnf.format( nruns) + " runs in " + lnf.format( runTimeNS) + " ns (" + lnf.format( ( double) runTimeNS / nruns) + " ns/call) with"
                            + "\n\t" + lnf.format( tc.numMismatchToFirst) + " different from first origin (" + lnf.format( 100.0 * tc.numMismatchToFirst / nruns) + "%)"
                            + "\n\t" + lnf.format( tc.numMismatchToLast) + " jumps from last origin (" + lnf.format( 100.0 * tc.numMismatchToLast / nruns) + "%)"
                            + "\n\t" + lnf.format( tc.numMismatchToFirstBig) + " different from first origin by more than " + BIG_OFFSET_MS + " ms"
                                    + " (" + lnf.format( 100.0 * tc.numMismatchToFirstBig / nruns) + "%)"
                            + "\n\t" + "total drift: " + lnf.format( originDrift) + " ms, " + lnf.format( skipped) + " skipped (" + lnf.format( 100.0 * skipped / nruns) + " %)");
                }};
            threads[ i].start();
        }
        try {
            for ( Thread thread : threads) {
                thread.join();
            }
        } catch ( InterruptedException ie) {};
    }

    public static long getNow( TimeCoherency coherency) {
        long    millisBefore = System.currentTimeMillis();
        long    now = System.nanoTime();
        if ( coherency != null) {
            checkOffset( now, millisBefore, coherency);
        }
        return now - startNanos;
    }

    private static void checkOffset( long nanoTime, long millisBefore, TimeCoherency tc) {
        long    millisAfter = System.currentTimeMillis();
        if ( millisBefore != millisAfter) {
            // disregard since thread may have slept between calls
            return;
        }
        tc.numChecks++;
        long    nanoMillis = ( long) ( nanoTime / 1e6);
        long    nanoOrigin = millisAfter - nanoMillis;
        long    oldOrigin = tc.lastOrigin;
        if ( oldOrigin != nanoOrigin) {
            tc.lastOrigin = nanoOrigin;
            tc.numMismatchToLast++;
        }
        if ( tc.firstOrigin != nanoOrigin) {
            tc.numMismatchToFirst++;
        }
        if ( Math.abs( tc.firstOrigin - nanoOrigin) > BIG_OFFSET_MS) {
            tc.numMismatchToFirstBig ++;
        }
    }
}

现在我做了一些小改动。基本上，我将 nanoTime 调用放在两个 currentTimeMillis 调用之间，以查看线程是否已被重新安排(这应该超过 currentTimeMillis 分辨率)。在这种情况下，我忽略循环周期。实际上，如果我们知道 nanoTime 足够快(如 Ivy Bridge 等较新的架构)，我们可以将 currentTimeMillis 与 nanoTime 括起来。

现在长 >10 毫秒的跳跃消失了。相反，我们会在距离每个线程的第一个起点超过 2 毫秒时进行计数。在我测试过的机器上，对于 100 秒的运行时间，调用之间总是有接近 200.000 次的跳转。对于这些情况，我认为 currentTimeMillis 或 nanoTime 可能不准确。

最佳答案

如前所述，每次计算新原点都意味着您容易出错。

//                               ______ delay _______
//                              v                    v
long origin = (long)(System.currentTimeMillis() - System.nanoTime() / 1e6);
//                                                                  ^
//                                                            truncation

如果您修改程序以便同时计算原点差异，您会发现它非常小。我测得大约 200ns 的平均值，这对于时间延迟来说是合适的。

使用乘法而不是除法(这应该没问题，再过几百年不会溢出)你还会发现计算出的未通过相等性检查的原点数量要大得多，大约 99%。如果错误的原因是因为时间延迟，则只有当延迟恰好与上一次相同时，它们才会通过。

一个更简单的测试是在对 nanoTime 的一些后续调用中累积耗时，并查看它是否检查了第一次和最后一次调用:

public class SimpleTimeCoherencyTest {
    public static void main(String[] args) {
        final long anchorNanos = System.nanoTime();

        long lastNanoTime = System.nanoTime();
        long accumulatedNanos = lastNanoTime - anchorNanos;

        long numCallsSinceAnchor = 1L;

        for(int i = 0; i < 100; i++) {
            TestRun testRun = new TestRun(accumulatedNanos, lastNanoTime);

            Thread t = new Thread(testRun);
            t.start();

            try {
                t.join();
            } catch(InterruptedException ie) {}

            lastNanoTime = testRun.lastNanoTime;
            accumulatedNanos = testRun.accumulatedNanos;
            numCallsSinceAnchor += testRun.numCallsToNanoTime;
        }

        System.out.println(numCallsSinceAnchor);
        System.out.println(accumulatedNanos);
        System.out.println(lastNanoTime - anchorNanos);
    }

    static class TestRun
    implements Runnable {
        volatile long accumulatedNanos;
        volatile long lastNanoTime;
        volatile long numCallsToNanoTime;

        TestRun(long acc, long last) {
            accumulatedNanos = acc;
            lastNanoTime = last;
        }

        @Override
        public void run() {
            long lastNanos = lastNanoTime;
            long currentNanos;

            do {
                currentNanos = System.nanoTime();
                accumulatedNanos += currentNanos - lastNanos;
                lastNanos = currentNanos;
                numCallsToNanoTime++;
            } while(currentNanos - lastNanoTime <= 100000000L);

            lastNanoTime = lastNanos;
        }
    }
}

该测试确实表明原点是相同的(或者至少错误是零均值)。

关于java - System.nanoTime 与 System.currentTimeMillis，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20874238/