根据其文档，System.nanoTime返回
自某个固定但任意的起始时间以来的十亿分之一秒。但是，在所有我尝试过以下代码的x64机器上，都发生了时间跳跃，从而移动了固定的原点时间。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。但是，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，我遇到了尝试测量延迟的问题，有时我会得到非常负面的延迟。在那些情况下，开始和结束时间戳记是由不同的线程创建的，但是延迟是在这些线程完成之后计算的。
我在这里的代码使用nanoTime花费时间，计算currentTimeMillis时间中的固定起点，并在两次调用之间比较该起点。既然我必须在这里提出一个问题:该代码有什么问题？为什么会看到违反固定来源契约(Contract)的情况？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，我做了一些小的更改。基本上，我将nanoTime调用放在两个currentTimeMillis调用之间，以查看线程是否已重新调度(应该比currentTimeMillis分辨率花费更多时间)。在这种情况下，我无视循环周期。实际上，如果我们知道nanoTime足够快(例如在Ivy Bridge等较新的体系结构上)，则可以将currentTimeMillis与nanoTime合并在一起。
现在，> 10ms的长跳转就消失了。相反，我们计算每个线程何时与第一个原点相距2ms以上。在我测试过的机器上，对于100秒的运行时间，两次调用之间总是有接近200.000的跳转。对于这些情况，我认为currentTimeMillis或nanoTime可能都不准确。
最佳答案
如前所述，每次计算一个新的原点都意味着您会出错。
// ______ delay _______
// v v
long origin = (long)(System.currentTimeMillis() - System.nanoTime() / 1e6);
// ^
// truncation
如果修改程序，以便同时计算原点差，您会发现它很小。我测得的平均时间约为200ns，这对于时间延迟来说是正确的。
使用乘法而不是除法(应该再好几百年而不会溢出)，您还会发现计算出的通过均等检查失败的原点数量要大得多，约为99％。如果错误的原因是由于时间延迟，则只有在延迟恰好与最后一个延迟相同时，它们才会通过。
一个更简单的测试是累积对nanoTime的一些后续调用所花费的时间，并查看是否检查出第一个和最后一个调用:
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(或者至少误差是零均值)。
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/