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：
- 从1开始打印整数
- 线程t1, 打印非5的倍数
- 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：
- 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...
- 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 data
std::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。
