---

semaphore buffer=1000,缓冲区剩余

semaphore goods = 1000-buffer

metux lock = 1

product{

p(buffer)

//生产

v(goods)

}

consumer{

int i = 0;

p（lock）

while{

if（i>=10）{

V(lock)

}

p(goods)

//消费

v(buffer)

}

}