我正在阅读Java The complete Reference - Herbert Schildt (TATA McGRAW HILL)中的主题。在这个示例表单中，优先级较低的Thread应该比具有较高优先级的Thread提供较低的clicks值。但我得到了什么
low-priority thread 3000255895
hi-priority thread 2857361716
根据书，它是这样的（差异）
low-priority thread 4408112
hi-priority thread 589626904
码
class Clicker implements Runnable {
long click = 0;
Thread t;
private volatile boolean running = true;
public Clicker (int p) {
t = new Thread(this);
t.setPriority(p);
}
public void run() {
while(running) {
click++;
}
}
public void stop() {
running = false;
}
public void start() {
t.start();
}
}
class Priority {
public static void main(String args[]) {
Thread.currentThread().setPriority(Thread.MAX_PRIORITY);
Clicker hi = new Clicker(Thread.NORM_PRIORITY + 2);
Clicker lo = new Clicker(Thread.NORM_PRIORITY - 2);
hi.start();
lo.start();
try {
Thread.sleep(10000);
} catch (InterruptedException e) {
System.out.println("main thread interrupted");
}
lo.stop();
hi.stop();
try {
hi.t.join();
lo.t.join();
} catch (InterruptedException e) {
System.out.println("interrupted exception catched in main");
}
System.out.println("low-priority thread " + lo.click);
System.out.println("hi-priority thread " + hi.click);
}
}
我在Intel® Core™ i3-2310M CPU @ 2.10GHz × 4 with Ubuntu 13.04
Thread.sleep(50000);
low-priority thread 14849893875
hi-priority thread 14224080358
这不是一个很大的区别
我也尝试优先级HI = 10和LOw = 1，但结果几乎相同。
更新：
运行100个线程50后，priority = 1和50 with priority = 10结果为
此处time = clicks values
high = 82 time = 110117529
low = 83 time = 102549208
high = 84 time = 110905795
low = 85 time = 99100530
high = 86 time = 105012756
low = 87 time = 110195297
high = 88 time = 102820088
low = 89 time = 97814606
high = 90 time = 99990839
low = 91 time = 102326356
high = 92 time = 98656119
low = 93 time = 98127243
high = 94 time = 97097823
low = 95 time = 103604394
high = 96 time = 93632744
low = 97 time = 99032550
high = 98 time = 103879116
low = 99 time = 97179029
使用1000个主题Take 4-5 min. to complete, all 4 cores at 100%
low = 977 time = 23593983
high = 978 time = 23998970
low = 979 time = 23879458
high = 980 time = 22775297
low = 981 time = 21297504
high = 982 time = 22464600
low = 983 time = 20301501
high = 984 time = 19073992
low = 985 time = 19450707
high = 986 time = 19317769
low = 987 time = 18454648
high = 988 time = 17901939
low = 989 time = 16976661
high = 990 time = 17360728
low = 991 time = 16813824
high = 992 time = 15531501
low = 993 time = 14659965
high = 994 time = 12833364
low = 995 time = 13708670
high = 996 time = 13348568
low = 997 time = 12947993
high = 998 time = 12749436
low = 999 time = 9804643
每个线程的不可预测的输出。
答案 0 :(得分：2)
我怀疑示例数字在某种程度上取决于您的平台和作者（示例是否运行在单个核心上？）。 JVM线程优先级是特定于OS /平台的。来自this interesting document
Linux优先级
在Linux下，你必须通过更多的箍来获取线程 最重要的是，尽管最后，它们可能更多 比在Windows下有用。在Linux中：
- 线程优先级仅在Java 6之后起作用;
- 要让它们工作，您必须以root身份运行（或通过setuid以root权限运行）;
- JVM参数-XX：必须包含UseThreadPriorities
请检查您的Java版本，Java调用和您的权限。我怀疑（！）你只满足其中一个标准（你的Java / JVM版本）。
答案 1 :(得分：0)
除了Thread优先级在大多数实现和平台下都是灰色区域之外，您可以查看的其他内容是：
很可能，click不会包含确切的调用次数。这应该是AtomicLong。那么你想要差异有多大？ + -2也没有太大的区别。
使用MIN_PRIORITY运行一个线程，使用MAX_PRIORITY运行其他线程。计算差异。用10,100,1000个线程重复上述实验。其中50％使用MIN_PRIORITY，其他50％使用MAX_PRIORITY。如果你发布你的结果肯定会启发
修改 有意思，让我提交我的示例程序。我在OpenJDK6和线程优先级下：MIN_PRIORITY和MAX_PRIORITY似乎并不重要。高级线程的点击次数略高于低优先级线程的点击次数。
import java.util.concurrent.atomic.AtomicLong;
class Clicker implements Runnable {
AtomicLong click = new AtomicLong();
Thread t;
private volatile boolean running = true;
public Clicker (int p) {
t = new Thread(this);
t.setPriority(p);
}
public void run() {
while(running) {
click.incrementAndGet();
longOP();
}
}
private void longOP() {
outer:
for (int i = 2; i <= 10000 && running; i++) {
final int NUM = i;
for (int j = 2; j * j <= NUM && running; j++) {
if (NUM % j == 0) {
continue outer;
}
}
System.out.print("[" + NUM + "],");
}
System.out.println();
for (int i = -90; i <= 90 && running; i++) {
double rad = Math.toRadians(i);
double value = Math.sqrt(Math.abs(Math.pow(Math.PI * Math.sin(rad), 10.0))) + Math.tan(rad) * Math.cos(rad); //lol
System.out.print("[" + i + " " + value + "],");
}
}
public void stop() {
running = false;
}
public void start() {
t.start();
}
}
public class Priority {
public static void main(String args[]) {
Thread.currentThread().setPriority(Thread.MAX_PRIORITY);
final int MAX_THREADS = 1000;
final long MAX_RUN_TIME = 60 * 1000; //millis
Clicker clickHi[] = new Clicker[MAX_THREADS];
Clicker clickLow[] = new Clicker[MAX_THREADS];
for (int i = 0; i < MAX_THREADS; ++i) {
clickHi[i] = new Clicker(Thread.MAX_PRIORITY);
clickLow[i] = new Clicker(Thread.MIN_PRIORITY);
clickHi[i].start();
clickLow[i].start();
}
try {
Thread.sleep(MAX_RUN_TIME);
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
System.out.println("main thread interrupted");
}
for (int i = 0; i < MAX_THREADS; ++i) {
clickHi[i].stop();
clickLow[i].stop();
}
for (int i = 0; i < MAX_THREADS; ++i) {
try {
clickHi[i].t.join();
clickLow[i].t.join();
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
System.out.println("interrupted exception catched in main");
}
}
long totalLowClicks = 0, totalHighClicks = 0;
for (int i = 0; i < MAX_THREADS; ++i) {
totalLowClicks += clickLow[i].click.longValue();
totalHighClicks += clickHi[i].click.longValue();
System.out.println("low-priority thread " + clickLow[i].click);
System.out.println("hi-priority thread " + clickHi[i].click);
}
System.out.println("Sum of LOW clicks : " + totalLowClicks);
System.out.println("Sum of HI clicks : " + totalHighClicks);
}
}
以上程序是cpu密集型的。您可能会遇到内存不足错误。但只是为了好玩，发布你的结果。