C ++ 0x没有信号量？如何同步线程？

问题描述

是不是C ++ 0x会没有信号量？在Stack Overflow上已经有一些关于信号量使用的问题。我使用它们（posix信号量）让一个线程等待另一个线程中的一些事件：

Is it true that C++0x will come without semaphores? There are already some questions on Stack Overflow regarding the use of semaphores. I use them (posix semaphores) all the time to let a thread wait for some event in another thread:

void thread0(...)
{
  doSomething0();

  event1.wait();

  ...
}

void thread1(...)
{
  doSomething1();

  event1.post();

  ...
}

用互斥体：

void thread0(...)
{
  doSomething0();

  event1.lock(); event1.unlock();

  ...
}

void thread1(...)
{
  event1.lock();

  doSomethingth1();

  event1.unlock();

  ...
}

问题：它不能保证thread1首先锁定互斥体（假定同一个线程应该锁定和解锁互斥体，你也不能在thread0和thread1开始之前锁定event1）。

Problem: It's ugly and it's not guaranteed that thread1 locks the mutex first (Given that the same thread should lock and unlock a mutex, you also can't lock event1 before thread0 and thread1 started).

因此，既然boost也没有信号量，那么最简单的方法是什么呢？

So since boost doesn't have semaphores either, what is the simplest way to achieve the above?

推荐答案

一个来自互斥体和条件变量：

You can easily build one from a mutex and a condition variable:

#include <boost/thread/condition.hpp>
#include <boost/thread/mutex.hpp>

class semaphore
{
private:
    boost::mutex mutex_;
    boost::condition_variable condition_;
    unsigned long count_;

public:
    semaphore()
        : count_()
    {}

    void notify()
    {
        boost::mutex::scoped_lock lock(mutex_);
        ++count_;
        condition_.notify_one();
    }

    void wait()
    {
        boost::mutex::scoped_lock lock(mutex_);
        while(!count_)
            condition_.wait(lock);
        --count_;
    }
};

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