下面这段介绍，修改自 wxWidgets 官方文档（详见：wxSemaphore Class Reference）。

semaphore is a counter limITing the number of threads concurrently accessing a shared resource.

This counter is always between 0 and the maximum value sPEcified during the SEMaphore creation. When the counter is strictly greater than 0, a call to Wait returns immediately and decrements the counter. As soon as it reaches 0, any subsequent calls to Semaphore::Wait block and only return when the semaphore counter becomes strictly positive again as the result of calling Signal which increments the counter.

In general, semaphores are useful to restrict access to a shared resource which can only be accessed by some fixed number of clients at the same time. For example, when modeling a hotel reservation System a semaphore with the counter equal to the total number of available rooms could be created. Each time a room is reserved, the semaphore should be acquired by calling Wait and each time a room is freed it should be released by calling Signal.

C++11 和 Boost.Thread 都没有提供信号量。对此 Boost 是这样解释的（Why has class semaphore disappeared?）：

Semaphore was removed as too error PRone. The same effect can be achieved with greater safety by the combination of a mutex and a condition VARiable. Dijkstra (the semaphore's inventor), Hoare, and brinch Hansen all depreciated semaphores and advocated more structured alternatives. In a 1969 letter to Brinch Hansen, Wirth said "semaphores ... are not suitable for higher level languages." [Andrews-83] summarizes typical errors as "omitting a P or a V, or accidentally coding a P on one semaphore and a V on on another", forgetting to include all references to shared objects in critical sections, and confusion caused by using the same primitive for "both condition synchronization and mutual exclusion".

简单来说，就是信号量太容易出错了（too error prone），通过组合互斥锁（mutex）和条件变量（condition variable）可以达到相同的效果，且更加安全。实现如下：

class Semaphore {
public:
  explicit Semaphore(int count = 0) : count_(count) {
  }

  void Signal() {
    std::unique_lock<std::mutex> lock(mutex_);
    ++count_;
    cv_.notify_one();
  }

  void Wait() {
    std::unique_lock<std::mutex> lock(mutex_);
    cv_.wait(lock, [=] { return count_ > 0; });
    --count_;
  }

private:
  std::mutex mutex_;
  std::condition_variable cv_;
  int count_;
};

下面创建三个工作线程（Worker），来测试这个信号量。

int main() {
  const std::size_t SIZE = 3;

  std::vector<std::thread> v;
  v.reserve(SIZE);

  for (std::size_t i = 0; i < SIZE; ++i) {
    v.emplace_back(&amp;Worker);
  }

  for (std::thread& t : v) {
    t.join();
  }
  
  return 0;
}

每个工作线程先等待信号量，然后输出线程 ID 和当前时间，输出操作以互斥锁同步以防止错位，睡眠一秒是为了模拟线程处理数据的耗时。

std::mutex g_io_mutex;

void Worker() {
  g_semaphore.Wait();

  std::thread::id thread_id = std::this_thread::get_id();

  std::string now = FormatTimeNow("%H:%M:%S");
  {
    std::lock_guard<std::mutex> lock(g_io_mutex);
    std::cout << "Thread " << thread_id << ": wait succeeded" << " (" << now << ")" << std::endl;
  }

  // Sleep 1 second to simulate data processing.
  std::this_thread::sleep_for(std::chrono::seconds(1));

  g_semaphore.Signal();
}

信号量本身是一个全局对象，count 为 1，一次只允许一个线程访问：

Semaphore g_semaphore(1);

输出为：

Thread 1d38: wait succeeded (13:10:10)
Thread 20f4: wait succeeded (13:10:11)
Thread 2348: wait succeeded (13:10:12)

可见每个线程相隔一秒，即一次只允许一个线程访问。如果把 count 改为 3：

Semaphore g_semaphore(3);

那么三个线程输出的时间应该一样：

Thread 19f8: wait succeeded (13:10:57)
Thread 2030: wait succeeded (13:10:57)
Thread 199c: wait succeeded (13:10:57)

最后附上 FormatTimeNow 函数的实现：

std::string FormatTimeNow(const char* format) {
  auto now = std::chrono::system_clock::now();
  std::time_t now_c = std::chrono::system_clock::to_time_t(now);
  std::tm* now_tm = std::localtime(&now_c);

  char buf[20];
  std::strftime(buf, sizeof(buf), format, now_tm);
  return std::string(buf);
}

参考：

• C++0x has no semaphores? How to synchronize threads?
• Semaphore (programming)