java中线程优先级是怎么回事给出一个例子

Posted 2021-01-08 mark-to-win

j6.线程优先级 （视频下载） （全部书籍）

java中线程的优先级用1-10之间的数字表示，数值越大优先级越高，默认的优先级为5。Java中的线程优先级是在Thread类中定义的常量 NORM_PRIORITY : 值为5，MAX_PRIORITY :值为10，MIN_PRIORITY : 值为1，缺省优先级为 NORM_PRIORITY。有关优先级的常用方法有两个：1）final void setPriority(int newp) : 修改线程的当前优先级  2）final int getPriority() : 返回线程的优先级。马克-to-win:线程的优先级不能担保线程的执行次序。优先级高的线程获取CPU执行的几率较大，优先级低的线程也有机会执行。参考http://www.javamex.com/tutorials/threads/priority_what.shtml其中有句：there‘d quite possibly be lower-priority threads that barely got any CPU at all, being continually starved by higher-priority threads that needed CPU. So Windows has a fallback mechanism, whereby a thread that hasn‘t run for a long time is given a temporary priority boost.大致意思：为防止低优先级的线程被饿死，Windows有个抗争的方法， 给低优先级的线程一个优先级的临时提升。。。。。。更多的请大家参考此网站。

例：1.6.1-本章源码

class ThreadMark_to_win extends Thread {
    static boolean go=true;
    private String s;
    int i;
    public ThreadMark_to_win(String s) {
        this.s = s;
    }
    public void run() {
        while(go){
            System.out.println(s+" 到了  "+i++);
            try {
                this.sleep(1);
            } catch (InterruptedException ex) {
                ex.printStackTrace();
            }
        }
    }
}

public class Test {
    public static void main(String[] args) {
        Thread t1 = new ThreadMark_to_win("线程1");
        Thread t2 = new ThreadMark_to_win("线程2");
        t1.setPriority(1);
        t2.setPriority(10);
        t1.start();
        t2.start();

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详情请进：http://www.mark-to-win.com/index.html?content=JavaBeginner/javaUrl.html&chapter=JavaBeginner/JavaBeginner6_web.html#HigherPriorityFirstWrong