多线程问题？？

Question

你应该让你的执行线程停顿一下，一边别的线程又执行的可能。
例如：
class myThreadone implements Runnable
{
public void run()
{
for(int i=0;i<100;i++){
    System.out.println(""+i);
                      try{
                          sleep(20); //20ms
                      }
                      catch(Exception e){
                      }
                  }
}
}

Answer 1

应该使用yield()方法来处理。class myThreadone implements Runnable
{
public void run()
{
for(int i=0;i<100;i++){
    System.out.println(""+i);
                      try{
                          yield(); //20ms
                      }
                      catch(Exception e){
                      }
                  }
}
}

Answer 2

class myThreadone implements Runnable
{
public void run()
{
for(int i=0;i<100;i++){
    System.out.println(""+i);
                      try{
                          Thread.yield();                       }
                      catch(Exception e){
                      }
                  }
}
}

Answer 3

都大意了,
sleep()与yield()皆是Thread的类方法,
怎么可以在实现Runnable接口的类中直接用呢.要想实现
0
100
1
101
2
102
.
.
.
98
198
99
199
的效果,你要
class myThreadone implements Runnable
{
public void run()
{
  for(int i=0;i<100;i++)
  {
    System.out.println(""+i);
    try{
               Thread. yield(); 
               }
             catch(Exception e){}   }
 }
}class myThreadtwo implements Runnable
{
public void run()
{
 for(int i=100;i<200;i++)
 {
            System.out.println(i); 
            try{
               Thread. yield();
               }
            catch(Exception e){}
          }
}
}

Answer 4

class Runnablex implements Runnable{
public void run(){
try{
for(int i=0;i<10;i++){
Thread.sleep(1000);
System.out.println("ThreadM");
 }
}
  catch(InterruptedException ex){
    ex.printStackTrace();
    }
  }
}

Answer 5

我执行了你的程序了，可以出现数字交错的现象啊
虽然不是一个一个的交错，但是确实能体现线程在cpu队列里排队的性质

Answer 6

public class TestThread { public static void main(String[] args) {

Thread  thread1=new Thread(new myThreadone());
        Thread thread2=new Thread(new myThreadtwo());
thread1.start();
thread2.start();
}
}
class myThreadone implements Runnable
{
public void run()
{
for(int i=0;i<100;i++)
                {
                 try{
                   System.out.println(""+i); 
               Thread.yield();
                
               }catch(Exception e){e.printStackTrace();}
                }
         
}
}
class myThreadtwo implements Runnable
{
public void run()
{         
for(int i=100;i<200;i++)
              {
              try{
                 System.out.println(i);  
             Thread.yield();
               
            }catch(Exception e){e.printStackTrace();}
              }
           
}
}
不用信号量了，你的线程不多，略微改动一下就可实现你需要的要求

Answer 7

可以看看java tutorial关于Thread类的讲解，有个一样的问题，说得很清楚

Answer 8

public class TestThread {        public static void main(String[] args) {                Thread  thread1=new Thread(new myThreadone());
                Thread thread2=new Thread(new myThreadtwo());
                thread1.start();
                thread2.start();
        }
}
class myThreadone implements Runnable
{
        public void run()
        {
                for(int i=0;i<100;i++){
                  System.out.println("" + i);
                  Thread.yield();
                }
        }
}
class myThreadtwo implements Runnable
{
        public void run()
        {
                for(int i=100;i<200;i++){
                  System.out.println(i);
                  Thread.yield() ;
                }
        }
}