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今天无意中看到一篇帖子，关于条件变量的，不过仔细看看发现它并达不到原本的目的。

程序如下，读者可以先想想他的本意，以及有没有问题：

#include <iostream>
#include <thread>
#include <condition_variable>
#include <mutex>
#include <unistd.h>
using namespace std;
//全局条件变量
condition_variable cond;
mutex _mutex;
int count = 0;void fun1(){while(1){count++;unique_lock<mutex>lock(_mutex);if(count%5 == 0){cond.notify_one();}else{cout<<"this is fun1,count="<<count<<endl;}lock.unlock();sleep(1);}
}void fun2()
{while(1){unique_lock<mutex>lock(_mutex);cond.wait(lock);cout<<"this is fun2,count="<<count<<endl;lock.unlock();sleep(2);}
}int main()
{thread t1(fun1);thread t2(fun2);t1.join();t2.join();return 0;
}

OK，本意显然是：

1. 从1开始打印整数
2. 线程t1, 打印非5的倍数
3. 线程t2, 打印5的倍数

编译执行，运行的还不错，符合预期，但这都是sleep的功劳。把fun1中的sleep去掉，fun2中的sleep放到cond.wait(lock)后，它BUG的面目就暴露出来了：

void fun1(){while(1){count++;unique_lock<mutex>lock(_mutex);if(count%5 == 0){cond.notify_one();}else{cout<<"this is fun1,count="<<count<<endl;}lock.unlock();}
}void fun2()
{while(1){unique_lock<mutex>lock(_mutex);cond.wait(lock);sleep(2);cout<<"this is fun2,count="<<count<<endl;lock.unlock();}
}

[mzhai@lock]$ ./a.out
this is fun1,count=1
this is fun1,count=2
this is fun1,count=3
this is fun1,count=4
this is fun2,count=6
this is fun1,count=6
this is fun1,count=7
this is fun1,count=8
this is fun1,count=9
this is fun1,count=11
this is fun1,count=12
this is fun1,count=13
this is fun1,count=14
this is fun1,count=16
this is fun1,count=17
this is fun1,count=18
this is fun1,count=19
this is fun1,count=21

多线程结果不能因随机加了几个sleep就不同，加sleep仅仅是模拟线程调度不大一样了。

再回过头来看看代码哪些地方有问题：

1. cond.notify_one(); count是5的倍数时，t1会通过notify_one通知t2做事，但并不会阻止t1继续执行。想想一下如果t1执行的很快而t2一直没得到调度，则t1会打印1，2，3，4，6，7，8，9，11...
2. cond.wait(lock); 可能会假唤醒，此时t1并没有通知它。

那“this is fun2,count=6” 是怎么回事哪？不应该是5吗？一种可能性是（可以通过GDB调试来模拟）：

说了那么多，怎么改哪？

 这是一个典型的你等我我等你的例子，对于这个例子都是一方干完事情另一方才能继续，完全串休化的任务，直接写到一个线程里即可。如果说我为了练习线程同步技巧非要整两个线程，那也行，condition_variable官方文档上就有一个例子实现了main线程等待worker_thread完成任务：

#include <condition_variable>
#include <iostream>
#include <mutex>
#include <string>
#include <thread>std::mutex m;
std::condition_variable cv;
std::string data;
bool ready = false;
bool processed = false;void worker_thread()
{// Wait until main() sends datastd::unique_lock lk(m);cv.wait(lk, []{ return ready; });// after the wait, we own the lock.std::cout << "Worker thread is processing data\n";data += " after processing";// Send data back to main()processed = true;std::cout << "Worker thread signals data processing completed\n";// Manual unlocking is done before notifying, to avoid waking up// the waiting thread only to block again (see notify_one for details)lk.unlock();cv.notify_one();
}int main()
{std::thread worker(worker_thread);data = "Example data";// send data to the worker thread{std::lock_guard lk(m);ready = true;std::cout << "main() signals data ready for processing\n";}cv.notify_one();// wait for the worker{std::unique_lock lk(m);cv.wait(lk, []{ return processed; });}std::cout << "Back in main(), data = " << data << '\n';worker.join();
}

我们依样画葫芦：

#include <iostream>
#include <thread>
#include <condition_variable>
#include <mutex>
#include <unistd.h>
using namespace std;
//全局条件变量
condition_variable cond;
mutex _mutex;
bool ready = false;
bool processed = false;int count = 0;void fun1(){while(1){count++;unique_lock<mutex> lock1(_mutex);if(count%5 == 0){ready = true;processed = false;lock1.unlock();cond.notify_one();lock1.lock();cond.wait(lock1, []{ return processed; });}else{cout<<"this is fun1,count="<<count<<endl;}lock1.unlock();}
}void fun2()
{while(1){unique_lock<mutex> lock1(_mutex);cond.wait(lock1, []{ return ready; });cout<<"this is fun2,count="<<count<<endl;processed = true;ready = false;lock1.unlock();cond.notify_one();}
}int main()
{thread t1(fun1);thread t2(fun2);t1.join();t2.join();return 0;
}

结果符合预期，感兴趣的读者可以到处插入sleep测试一下。