根据CPU的核数和计算能力设置
但是一个java程序启动本身就会启动很多线程,帮我们分配资源的线程,还有操作系统的线程等
出于安全的角度,还需要给CPU留有20%的余地
实际需要压测,得到一个合理的数据。
当然也可以根据公式说明:
Nthreads = NcPu * Ucpu *(1 + W/C)其中:
Ncpu是处理器的核的数目,可以通过Runtime.getRuntime().availableProcessors()得到
Ucpu是期望的CPU利用率( 该值应该介于0和1之间)
具体设置多少,也可以根据实际运行的情况。下面代码展示:
package com.zy.c_000_threadbasic;
import java.text.DecimalFormat;
import java.util.Random;
import java.util.concurrent.CountDownLatch;
// 验证是否创建线程数越多越好
public class To1_MultiVSSingle_ContextSwitch {
private static double[] nums = new double[100000000];
private static Random random = new Random();
private static DecimalFormat df = new DecimalFormat("0.00");
static {
for (int i = 0; i < nums.length; i++) {
// 用于返回一个大于或等于 0.0 且小于 1.0 的随机浮点数
nums[i] = random.nextDouble();
}
}
public static void main(String[] args) throws Exception {
m1();
m2();
m3();
}
// 单线程运行
private static void m1() {
long start = System.currentTimeMillis();
double result = 0.0;
for (int i = 0; i < nums.length; i++) {
result += nums[i];
}
long end = System.currentTimeMillis();
System.out.println("m1:" + (end - start) + " result= " + df.format(result));
}
// 使用多线程的思维(2个)
static double result1 = 0.0, result2 = 0.0, result = 0.0;
private static void m2() throws InterruptedException {
Thread t1 = new Thread(() -> {
for (int i = 0; i < nums.length / 2; i++) {
result1 += nums[i];
}
});
Thread t2 = new Thread(() -> {
for (int i = nums.length / 2; i < nums.length; i++) {
result2 += nums[i];
}
});
long start = System.currentTimeMillis();
t1.start();
t2.start();
t1.join();
t2.join();
result = result1 + result2;
long end = System.currentTimeMillis();
System.out.println("m2:" + (end - start) + " result= " + df.format(result));
}
// 使用多线程的思维(多个)
private static void m3() throws InterruptedException {
final int threadCount = 32;
Thread[] threads = new Thread[threadCount];
double[] results = new double[threadCount];
final int segmentCount = nums.length / threadCount;
// 初始化,指定线程个数
CountDownLatch latch = new CountDownLatch(threadCount);
for (int i = 0; i < threadCount; i++) {
int m = i;
threads[i] = new Thread(() -> {
for (int j = m * segmentCount; j < (m + 1) * segmentCount && j < nums.length; j++) {
results[m] += nums[j];
}
// 每个线程执行后执行latch.countDown();,代表一个线程执行完成,待完成的线程数减1。
latch.countDown();
});
}
double resultM3 = 0.0;
long start = System.currentTimeMillis();
for (Thread t : threads) {
t.start();
}
// 阻塞该线程,等待其他子线程完成。
latch.await();
for (int i = 0; i < results.length; i++) {
resultM3 += results[i];
}
long end = System.currentTimeMillis();
System.out.println("m3:" + (end - start) + " result= " + df.format(resultM3));
}
}