多线程的通信问题，麻烦大家帮我看看
如果加上try{Thread.sleep(1);}catch(Exception w){}（24行-27行）要求的打印的效果没问题
张宇，男
李梅，女
张宇，男
李梅，女
但是没有上面那个sleep就不行了
这是为什么呀！！！？？？请各位帮帮忙~~~谢谢
class Q
{
String name = "unknown ";
String sex = "unknown ";
boolean bFull = false;


}
class producer implements Runnable
{
Q q = null;

public producer(Q q)
{
this.q = q;
}
public void run()
{
int i =0 ;
while(true)
{

synchronized(q)
{
if(q.bFull)
try{wait();}
catch(Exception w){}
if(i==0)
{
try{Thread.sleep(1);}
catch(Exception w){}
q.name = "张宇 ";
q.sex = "男 ";
}
else
{
q.name = "李梅 ";
q.sex = "女 ";
}
i = (i+1)%2;
q.bFull = true;
q.notify();
}

}
}
}
class customer implements Runnable
{
Q q = null;
//int j = 0;
public customer(Q q)
{
this.q = q;
}
public void run()
{

while(true)
{
synchronized(q)
{
if(!q.bFull)
try{wait();}
catch(Exception w){}
System.out.println(q.name+ "-----> " + q.sex);
q.bFull = false;
q.notify();
}
}
}
}

class TestThread
{
public static void main(String [] args)
{
Q q = new Q();
new Thread(new producer(q)).start();
new Thread(new customer(q)).start();
}
}

[解决办法]
作为生产者和消费者用一个对象做队列肯定是不对的。因为q.notify();有可能唤醒的是生产者也可能是消费者
class Q
{
String name = "unknown ";
String sex = "unknown ";
boolean bFull = false;
}
class ProducterQ{
Q q;
public ProducterQ(Q q){
this.q=q;
}
}
class CustomerQ{
Q q;
public CustomerQ(Q q){
this.q=q;
}
}
class producer implements Runnable
{
ProducterQ pq = null;
CustomerQ cq=null;

public producer(ProducterQ pq,CustomerQ cq)
{
this.pq = pq;
this.cq = cq;
}
public void run()
{
int i =0 ;
while(true)
{

synchronized(pq)
{
while(pq.q.bFull){
try{
pq.wait();
}
catch(Exception w){}
}
if(i==0)
{
pq.q.name = "张宇 ";
pq.q.sex = "男 ";
}
else
{
pq.q.name = "李梅 ";
pq.q.sex = "女 ";
}
i = (i+1)%2;
pq.q.bFull = true;

}
synchronized(cq)
{
cq.notify();

}

}
}
}
class customer implements Runnable
{
CustomerQ cq = null;
ProducterQ pq = null;
//int j = 0;
public customer(ProducterQ pq,CustomerQ cq)
{
this.cq = cq;
this.pq = pq;
}
public void run()
{

while(true)
{

synchronized(cq)
{
while(!cq.q.bFull){
try{
cq.wait();
}
catch(Exception w){}
}
System.out.println(cq.q.name+ "-----> " + cq.q.sex);
cq.q.bFull = false;

}
synchronized(pq)
{
pq.notify();
}
}
}
}

public class TestThread
{
public static void main(String [] args)
{
Q q = new Q();
ProducterQ pq=new ProducterQ(q);
CustomerQ cq=new CustomerQ(q);
new Thread(new producer(pq,cq)).start();
new Thread(new customer(pq,cq)).start();
}
}
[解决办法]
only issue of the lock , don 't need create ProducterQ and CustomerQ .

public class TestThread
{
public static void main(String [] args)
{
Q q = new Q();
Object lock1 =new Object();
Object lock2 =new Object();
new Thread(new producer(q,lock1,lock2)).start();
new Thread(new customer(q,lock1,lock2)).start();
}
}
[解决办法]
notify()
方法不能确定唤醒的是哪个线程
按官方的说法是唤醒顺序完全没有规则....
不按优先级,也不按前后顺序...