c++ - 在 Boost 中实现自旋锁。需要的例子

Question

我想知道 boost 是否有任何库可以帮助实现自旋锁。我知道 boost 支持互斥锁，但我找不到任何显示或描述 boost 中的自旋锁的示例。任何显示如何使用 boost（最好）实现自旋锁的示例将不胜感激。（C++98）

Answer 1

使用示例Boost.Atomic：

#include <boost/atomic.hpp>

class SpinLock
{
    boost::atomic_flag flag; // it differs from std::atomic_flag a bit -
                             // does not require ATOMIC_FLAG_INIT
public:
    void lock()
    {
        while( flag.test_and_set(boost::memory_order_acquire) )
            ;
    }
    bool try_lock()
    {
        return !flag.test_and_set(boost::memory_order_acquire);
    }
    void unlock()
    {
        flag.clear(boost::memory_order_release);
    }
};

Coliru 上的现场演示

#include <boost/range/algorithm.hpp>
#include <boost/atomic.hpp>
#include <boost/thread.hpp>
#include <iostream>
#include <vector>

class SpinLock
{
    boost::atomic_flag flag;
public:
    void lock()
    {
        while( flag.test_and_set(boost::memory_order_acquire) )
            ;
    }
    bool try_lock()
    {
        return !flag.test_and_set(boost::memory_order_acquire);
    }
    void unlock()
    {
        flag.clear(boost::memory_order_release);
    }
};

int main()
{
    using namespace std; using namespace boost;

    SpinLock lock;
    vector<thread> v;
    for(auto i = 0; i!=4; ++i)
        v.emplace_back([&lock, i]
        {
            for(auto j = 0; j!=16; ++j)
            {
                this_thread::yield();
                lock_guard<SpinLock> x(lock);
                cout << "Hello from " << i << flush << "\tj = " << j << endl;
            }
        });
    for(auto &t: v)
        t.join();
}

输出是：

Hello from 0    j = 0
Hello from 1    j = 0
Hello from 3    j = 0
Hello from 2    j = 0
Hello from 3    j = 1
Hello from 1    j = 1
Hello from 3    j = 2
Hello from 2    j = 1
Hello from 1    j = 2
Hello from 2    j = 2
Hello from 1    j = 3
Hello from 2    j = 3
Hello from 1    j = 4
Hello from 3    j = 3
Hello from 2    j = 4
Hello from 1    j = 5
Hello from 2    j = 5
Hello from 1    j = 6
Hello from 2    j = 6
Hello from 1    j = 7
Hello from 2    j = 7
Hello from 1    j = 8
Hello from 2    j = 8
Hello from 3    j = 4
Hello from 2    j = 9
Hello from 3    j = 5
Hello from 1    j = 9
Hello from 2    j = 10
Hello from 1    j = 10
Hello from 2    j = 11
Hello from 3    j = 6
Hello from 1    j = 11
Hello from 2    j = 12
Hello from 3    j = 7
Hello from 1    j = 12
Hello from 2    j = 13
Hello from 3    j = 8
Hello from 2    j = 14
Hello from 3    j = 9
Hello from 1    j = 13
Hello from 2    j = 15
Hello from 1    j = 14
Hello from 3    j = 10
Hello from 1    j = 15
Hello from 3    j = 11
Hello from 3    j = 12
Hello from 3    j = 13
Hello from 3    j = 14
Hello from 3    j = 15
Hello from 0    j = 1
Hello from 0    j = 2
Hello from 0    j = 3
Hello from 0    j = 4
Hello from 0    j = 5
Hello from 0    j = 6
Hello from 0    j = 7
Hello from 0    j = 8
Hello from 0    j = 9
Hello from 0    j = 10
Hello from 0    j = 11
Hello from 0    j = 12
Hello from 0    j = 13
Hello from 0    j = 14
Hello from 0    j = 15

Answer 2

这是一个使用 C++11 atomic 的示例：

#include <atomic>

typedef std::atomic<bool> Lock;

void enterCritical(Lock& lock) {
    bool unlocked = false;
    while (!lock.compare_exchange_weak(unlocked, true));
}

void exitCritical(Lock& lock) {
    lock.store(false);
}

Answer 3

The spinlock solution provided by erenon sometimes generated crumble cout result. but the boost::mutext solution won't. So either the solution is incorrect, or my understanding of cout is wrong.

#include <iostream>
#include <thread>
#include <atomic>
#include <boost/thread/mutex.hpp>
using namespace std;

class spinlock {
    private:
        std::atomic<bool> lock_;

    public:
        spinlock() {
            lock_.store(false);
        }

        void lock() {
            bool unlocked = false;
            while (!lock_.compare_exchange_weak(unlocked, true));

        }

        void unlock() {
            lock_.store(false);
        }
    };


class add_one
{
    private:
        std::string name_;
        unsigned int& num_;
        spinlock & lock_;
        boost::mutex & mutex_;
    public:
    add_one(std::string name, unsigned int& num, spinlock &lock, boost::mutex &mutex)
    :name_(name),
    num_(num),
    lock_(lock),
    mutex_(mutex)
    {

    }

    void add_and_display()
    {
        while(true)
        {
            lock_.lock();
            //boost::lock_guard<boost::mutex> g( mutex_);
            std::cout << name_ << " " << num_ << endl;
            if(num_ == 10000000)
            {

                return;
            }
            num_++;
            lock_.unlock();

        }
    }
};

int main()
{
   cout << "Hello World" << endl;
   unsigned int n = 0;
   spinlock lock;
   boost::mutex mutex;
   add_one one("t1", n ,lock, mutex);
   add_one two("t2", n ,lock, mutex);
   add_one three("t3", n ,lock, mutex);
   //add_one four("t4", n ,lock, mutex);
   std::thread t1(std::bind(&add_one::add_and_display, one));
   std::thread t2(std::bind(&add_one::add_and_display, two));
   std::thread t3(std::bind(&add_one::add_and_display, three));
   //std::thread t4(std::bind(&add_one::add_and_display, four));

   t1.join();
   t2.join();
   t3.join();
   //t4.join();

   return 0;
}

Answer 4

稍微偏离主题但相关——可以使用 C++11（而不是 C++03）获得无锁保证自旋锁，而无需boost（尽管那是问题标题）。

使用 C++11 std::atomic_flag。

正如 Evgeny Panasyuk 在erenon's answer的评论中指出的那样，它始终是无锁的。std::atomic似乎不能保证无需等待（在某些情况下需要）。

也符合BasicLockable接口，可以使用std::lock_guard// std::scoped_lock...

class Spinlock {
  std::atomic_flag _lock = ATOMIC_FLAG_INIT;

 public:
  void lock() {
    while (_lock.test_and_set(std::memory_order_acquire)) continue;
  }
  void unlock() {
    _lock.clear(std::memory_order_release);
  }
};