Ral*_*f H 7 java multithreading nanotime
根据其文档,System.nanoTime返回 纳秒,因为一些固定但任意的原始时间.但是,在所有x64机器上我尝试了下面的代码,有时间跳转,移动固定的原始时间.我的方法可能存在一些缺陷,使用替代方法(此处为currentTimeMillis)获取正确的时间.但是,测量相对时间(持续时间)的主要目的也会受到负面影响.
我在比较不同队列到LMAX的Disruptor时试图测量延迟时遇到了这个问题,我有时会出现非常负的延迟.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.
我的代码在这里花费时间使用nanoTime,计算currentTimeMillis时间中的固定原点,并比较调用之间的原点.因为我必须在这里提出一个问题:这段代码有什么问题？为什么会发现违反固定原产地合同的行为？或者不是吗？
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调用括在两个currentTimeMillis调用之间,以查看该线程是否已被重新调度(这应该超过currentTimeMillis分辨率).在这种情况下,我忽略了循环周期.实际上,如果我们知道nanoTime足够快(就像像Ivy Bridge这样的新架构),我们可以将currentTimeMillis括在nanoTime中.
现在长> 10ms的跳跃消失了.相反,我们计算每个线程距离第一个原点超过2毫秒的时间.在我测试过的机器上,对于100s的运行时间,调用之间总会有接近200,000次跳转.对于那些我认为currentTimeMillis或nanoTime可能不准确的情况.
正如已经提到的，每次计算一个新的原点意味着您可能会出错。
// ______ delay _______
// v v
long origin = (long)(System.currentTimeMillis() - System.nanoTime() / 1e6);
// ^
// truncation
如果您修改程序来计算原点差异，您会发现它非常小。我测得平均约为 200 纳秒，这对于时间延迟来说大约是正确的。
使用乘法而不是除法（这应该可以在接下来的几百年内不会溢出），您还会发现计算出的未通过相等性检查的起源数量要大得多，大约为 99%。如果错误的原因是由于时间延迟，则只有当延迟恰好与上一个延迟相同时，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经过的时间，并查看它是否与第一次和最后一次调用进行检查：
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）。