这里是一个微基准测试的例子:
public class ThreadSpawningPerformanceTest {
static long test(final int threadCount, final int workAmountPerThread) throws InterruptedException {
Thread[] tt = new Thread[threadCount];
final int[] aa = new int[tt.length];
System.out.print("Creating "+tt.length+" Thread objects... ");
long t0 = System.nanoTime(), t00 = t0;
for (int i = 0; i < tt.length; i++) {
final int j = i;
tt[i] = new Thread() {
public void run() {
int k = j;
for (int l = 0; l < workAmountPerThread; l++) {
k += k*k+l;
}
aa[j] = k;
}
};
}
System.out.println(" Done in "+(System.nanoTime()-t0)*1E-6+" ms.");
System.out.print("Starting "+tt.length+" threads with "+workAmountPerThread+" steps of work per thread... ");
t0 = System.nanoTime();
for (int i = 0; i < tt.length; i++) {
tt[i].start();
}
System.out.println(" Done in "+(System.nanoTime()-t0)*1E-6+" ms.");
System.out.print("Joining "+tt.length+" threads... ");
t0 = System.nanoTime();
for (int i = 0; i < tt.length; i++) {
tt[i].join();
}
System.out.println(" Done in "+(System.nanoTime()-t0)*1E-6+" ms.");
long totalTime = System.nanoTime()-t00;
int checkSum = 0;
for (int a : aa) {
checkSum += a;
}
System.out.println("Checksum: "+checkSum);
System.out.println("Total time: "+totalTime*1E-6+" ms");
System.out.println();
return totalTime;
}
public static void main(String[] kr) throws InterruptedException {
int workAmount = 100000000;
int[] threadCount = new int[]{1, 2, 10, 100, 1000, 10000, 100000};
int trialCount = 2;
long[][] time = new long[threadCount.length][trialCount];
for (int j = 0; j < trialCount; j++) {
for (int i = 0; i < threadCount.length; i++) {
time[i][j] = test(threadCount[i], workAmount/threadCount[i]);
}
}
System.out.print("Number of threads ");
for (long t : threadCount) {
System.out.print("\t"+t);
}
System.out.println();
for (int j = 0; j < trialCount; j++) {
System.out.print((j+1)+". trial time (ms)");
for (int i = 0; i < threadCount.length; i++) {
System.out.print("\t"+Math.round(time[i][j]*1E-6));
}
System.out.println();
}
}
}
在64位Windows 7上,使用32位Sun的Java 1.6.0_21客户端VM和Intel Core2 Duo E6400 @ 2.13 GHz处理器的结果如下:
Number of threads 1 2 10 100 1000 10000 100000
1. trial time (ms) 346 181 179 191 286 1229 11308
2. trial time (ms) 346 181 187 189 281 1224 10651
结论:如预期,两个线程的工作速度几乎是一个线程的两倍,因为我的计算机有两个核心。我的计算机每秒可以产生近10000个线程,即
线程创建开销为0.1毫秒。因此,在这样一台机器上,每秒产生几百个新线程对性能影响可以忽略不计(这也可以通过比较2个和100个线程列中的数字来看出)。
InheritableThreadLocal
协同工作。如果有人想尝试基于早期访问的 Java 17 的实验构建版本,则 Loom 团队正在征求反馈。 - Basil Bourque