这是一个微基准测试示例：
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; //display checksum in order to give the JVM no chance to optimize out the contents of the run() method and possibly even thread creation 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Intel Core2 Duo E6400 @ 2.13 GHz上使用32位Sun的Java 1.6.0_21 Client VM的64位Windows
7上的结果如下：
Number of threads 1 2 10 100 1000 10000 1000001. trial time (ms) 346 181 179 191 286 1229 113082. trial time (ms) 346 181 187 189 281 1224 10651
结论：由于我的计算机有两个内核，因此两个线程的工作速度几乎是预期的两倍。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，在这样的机器上，每秒几百个新线程造成的开销可以忽略不计（也可以通过比较2线程和100线程列中的数字来看出）。
