大虾们，为什么这个程序不能运行（线程方面的）？如何修改啊？
import   java.io.*;

class   StringRunnable   implements   Runnable
{
    public   static   String   str;
    public   StringRunnable()
    {
        System.out.println( "StringRunnable ");
    }

    public   void   run()
    {
try
{
            BufferedReader   in   =   new   BufferedReader(new   InputStreamReader(System.in));
            str   =   in.readLine();
            System.out.println(str);
        }
        catch   (IOException   e)
        {
    System.out.println(e.getMessage());
}
    }
}

public   class   StringThread   extends   Thread
{
    public   StringThread()
    {
        System.out.println( "StringThread ");
    }

    public   void   run()
    {
        System.out.println( "The   length   of   string   is:   "   +   StringRunnable.str.length());
    }

    public   static   void   main(String[]   args)
    {
        Thread   thread1   =   new   Thread(new   StringRunnable());
        thread1.setPriority(Thread.MAX_PRIORITY);
        thread1.start();
        Thread   thread2   =   new   StringThread();
        thread2.start();
    }
}

------解决方案--------------------
thread1.setPriority(Thread.MAX_PRIORITY); 
thread1优先级高，并不代表thread1不执行完thread2就不会被执行
当thread1走到这里
BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); 
str = in.readLine(); 
等到用户输入的时候，如果用户没有及时输入或者没有输入完成，thread1就会堵塞在这里

这时候，如果thread2开始了，运行到
System.out.println("The length of string is: " + StringRunnable.str.length()); 

由于StringRunnable.str还是null，调用length()就会造成thread2异常终止

如果LZ想让线程顺序执行，可以不用设优先级，直接用join就可以了

public static void main(String[] args) 
{ 
Thread thread1 = new Thread(new StringRunnable()); 
//thread1.setPriority(Thread.MAX_PRIORITY); 
//thread1.start(); 
Thread thread2 = new StringThread(); 
//thread2.start(); 
thread1.start();
thread1.jion();
thread2.start();
}