Question

我正在使用boost 1.54.0和Visual Studio 2010.对于代码：

#include <iostream>
#include "boost/thread/thread.hpp"
#include "boost/thread/mutex.hpp"

boost::mutex mx1;

void func1()
{
    {
        boost::mutex::scoped_lock(mx1);
        std::cout << "Thread " << boost::this_thread::get_id() << " starting work." << std::endl;
    }
    int x = 0;
    for (int i=0; i<100; i++)
        x++;
    {
        boost::mutex::scoped_lock(mx1);
        std::cout << "Thread " << boost::this_thread::get_id() << " finished." << std::endl;
    }
}

int main(void)
{
    boost::thread thread1(&func1);
    boost::thread thread2(&func1);
    thread1.join();
    thread2.join();
    return 0;
}

大约一半时间我得到以下内容（显然有不同的线程ID和执行顺序）：

Thread Thread 15b0 starting work.
1a18 starting work.
Thread 15b0 finished.
Thread 1a18 finished.

......而不是这个（这是我所期待的）：

Thread 15b0 starting work.
Thread 1a18 starting work.
Thread 15b0 finished.
Thread 1a18 finished.

然而，使用

mx1.lock();
std::cout << "Thread " << boost::this_thread::get_id() << " starting work." << std::endl;
mx1.unlock();

......似乎没有任何问题。

输出似乎总是遵循相同的模式。我是否错误地使用了互斥锁，还是与std :: cout有关？

Answer 1

替换

    boost::mutex::scoped_lock(mx1);

与

    boost::mutex::scoped_lock lock(mx1);

你成了最常发生错字的受害者： - ）

boost :: scoped_lock似乎没有锁定std :: cout

1 个答案: