java - yield 在这里如何运作？

Question

方法Thread.yield:

Causes the currently executing thread object to temporarily pause and allow other threads to execute.

所以在下面的代码中:

public class Test implements Runnable {  

    private int stopValue;  

    public Fib(int stopValue){  
        this.stopValue = stopValue;  
    }  

    @Override  
    public void run() {  

        System.out.println("In test thread");  
        for(int i = 0; i < stopValue; i++){  
            c = i + 1;  
        }  

        System.out.println("Result = "+c);        
    }  

    public static void main(String[] args){  
        int defaultStop = 1024;  
        if(args.length > 0){   
            defaultStop = Integer.parseInt(args[0]);  
        }  
        Thread a = new Thread(new Fib(defaultStop));  
        System.out.println("In main");  
        a.setDaemon(true);  
        a.start();  
        Thread.yield();       
        System.out.println("Back in main");  
    }  

}

我希望我应该看到:

1. 在 main 然后
2. 在测试线程中

其余的将是未定义的。但我不明白为什么有时我只看到:In main 和 Back in main 而不是来自 Test 线程的任何打印语句？

Answer 1

yield() 是对操作系统调度的提示，但不提供调度方面的任何保证。它并不总是停顿很长时间。如果您重复调用它，它可能只需要几微秒。

启动线程需要时间，即使您的主线程短暂暂停，它也可能在后台线程启动之前完成。

---

如您所见，yield() 暂停非常短暂。

long start = System.nanoTime();
long runs = 20000000;
for (int i = 0; i < runs; i++)
    Thread.yield();
long time = System.nanoTime() - start;
System.out.printf("Thread.yield() took an average of %,d ns.%n", time / runs);

打印

Thread.yield() took an average of 148 ns.

相比之下，System.nanoTime 在我的机器上花费的时间更长。

long start = System.nanoTime();
long runs = 20000000;
for (int i = 0; i < runs; i++)
    System.nanoTime();
long time = System.nanoTime() - start;
System.out.printf("System.nanoTime() took an average of %,d ns.%n", time / runs);

打印

System.nanoTime() took an average of 656 ns.

这两个时间因操作系统和机器而异。