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Is there an easy way to implement AutoResetEvent in C++0x?

I understand I've asked this question before: What is the C++ equivalent for AutoResetEvent under Linux?

However, I'm learning that in C++0x, the threading library are made much simpler, so I want to pose this question out again, is there an easy way to implement AutoResetEvent in C++0x?

like image 942
derekhh Avatar asked Dec 16 '11 18:12

derekhh


1 Answers

Here is a translation of the accepted answer to your first question to use C++11 tools:

#include <mutex>
#include <condition_variable>
#include <thread>
#include <stdio.h>

class AutoResetEvent
{
  public:
  explicit AutoResetEvent(bool initial = false);

  void Set();
  void Reset();

  bool WaitOne();

  private:
  AutoResetEvent(const AutoResetEvent&);
  AutoResetEvent& operator=(const AutoResetEvent&); // non-copyable
  bool flag_;
  std::mutex protect_;
  std::condition_variable signal_;
};

AutoResetEvent::AutoResetEvent(bool initial)
: flag_(initial)
{
}

void AutoResetEvent::Set()
{
  std::lock_guard<std::mutex> _(protect_);
  flag_ = true;
  signal_.notify_one();
}

void AutoResetEvent::Reset()
{
  std::lock_guard<std::mutex> _(protect_);
  flag_ = false;
}

bool AutoResetEvent::WaitOne()
{
  std::unique_lock<std::mutex> lk(protect_);
  while( !flag_ ) // prevent spurious wakeups from doing harm
    signal_.wait(lk);
  flag_ = false; // waiting resets the flag
  return true;
}


AutoResetEvent event;

void otherthread()
{
  event.WaitOne();
  printf("Hello from other thread!\n");
}


int main()
{
  std::thread h(otherthread);
  printf("Hello from the first thread\n");
  event.Set();

  h.join();
}

Output:

Hello from the first thread
Hello from other thread!

Update

In the comments below tobsen notes that AutoResetEvent has the semantics of signal_.notify_all() instead of signal_.notify_one(). I haven't changed the code because the accepted answer to the first question used pthread_cond_signal as opposed to pthread_cond_broadcast and I am leading with the statement that this is a faithful translation of that answer.

like image 108
Howard Hinnant Avatar answered Nov 11 '22 09:11

Howard Hinnant