我有两个线程类“AddThread”和“ReadThread”。这些线程的执行应如下所示:“AddThread应该添加1条记录，并等待ReadThread显示该记录，之后ReadThread应该再次显示该添加的记录AddThread应该添加另一条记录”，此过程应该继续进行，直到添加了所有记录(REcords为从LinkedList访问)。这是代码
class AddThread extends Thread
{
private Xml_Parse xParse;
LinkedList commonlist;
AddThread(LinkedList commonEmpList)
{
commonlist = commonEmpList;
}
public void run()
{
System.out.println("RUN");
xParse=new Xml_Parse();
LinkedList newList=xParse.xmlParse();
try
{
synchronized (this) {
if(newList.size()>0)
{
for(int i=0;i<newList.size();i++)
{
System.out.println("FOR");
commonlist.add(newList.get(i));
System.out.println("Added" +(i+1)+ "Record");
}
System.out.println(commonlist.size());
}
}
}
catch(Exception e)
{
}
}
}
class ReadThread extends Thread
{
LinkedList commonlist;
ReadThread(LinkedList commonEmpList)
{
commonlist = commonEmpList;
}
public void run()
{
try
{
synchronized (this) {
System.out.println();
System.out.println("ReadThread RUN");
sleep(1000);
//System.out.println("After waiting ReadThread RUN");
System.out.println(commonlist.size());
if(commonlist.size()>0)
{
for(int j=0;j<commonlist.size();j++)
{
System.out.println("Read For");
System.out.println("EmpNo: "+((EmployeeList)commonlist.get(j)).getEmpno());
System.out.println("EmpName: "+((EmployeeList)commonlist.get(j)).getEname());
System.out.println("EmpSal: "+((EmployeeList)commonlist.get(j)).getEmpsal());
}
}
}
}
catch(Exception e)
{
}
}
}
public class MainThread
{
public static LinkedList commonlist=new LinkedList();
public static void main(String args[])
{
AddThread addThread=new AddThread(commonlist);
ReadThread readThread=new ReadThread(commonlist);
addThread.start();
readThread.start();
}
}
最佳答案
您将需要学习如何有效使用 wait() 和 notify() 。
也可以看看: