boost::atomic 实现 spinlock

Enic — Sun, 31 Mar 2013 13:49:00 GMT

先看下boost�l�的例子�Q�我觉得有问题：

#include <boost/atomic.hpp>

class spinlock {
private:
  typedef enum {Locked, Unlocked} LockState;
  boost::atomic<LockState> state_;

public:
  spinlock() : state_(Unlocked) {}

  void lock()
  {
    // 可能�q�里的boost::memory_order_acquire有原子操作的效果吧，偶不是很理解�Q�不�q�我觉得�q�里应该用cae操作才对
    while (state_.exchange(Locked, boost::memory_order_acquire) == Locked) {
      /* busy-wait */
    }
  }
  void unlock()
  {
    // �q�里都直接写不做���查了�Q�更加难以理�?
    state_.store(Unlocked, boost::memory_order_release);
  }
};

有可能是我不理解后面的内存访问参数的意义�Q�给下我自己的实玎ͼ�
class CESpinLock : boost::noncopyable
{
private:
    typedef enum {emUnlocked = 0, emLocked} EM_LockState;

public:
    CESpinLock() : m_atomicState(emLocked)
    {
    }

public:
    void lock()
    {
        EM_LockState state = emUnlocked;
        while(false == m_atomicState.compare_exchange_strong(state, emLocked))
        {
            state = emUnlocked;
        }
    }

    void unlock()
    {
        EM_LockState state = emLocked;
        while(false == m_atomicState.compare_exchange_strong(state, emUnlocked))
        {
            state = emLocked;
        }
    }

private:
    boost::atomic<EM_LockState> m_atomicState;
};

可以适当的在false里边加一点sleep操作感觉�?br />
�q�有一点就是不太激烈这里的cae操作分两�U?strong和weak

bool compare_exchange_weak(T & expected, T desired, memory_order success_order, memory_order failure_order)

bool compare_exchange_strong(T & expected, T desired, memory_order order)

Compare current value with expected, change it to desired if matches. Returns true if an exchange has been performed, and always writes the previous value back in expected.

实在不理�?nbsp;May fail spuriously, so must generally be retried in a loop.的意义，不过看了代码�Q�在win32的实��C��Q�weak是调用了stong实现的�?br />

VCZH.�_�丝数组[0]  21:49:07
atomic�?nbsp;compare_exchange_weak
compare_exchange_weak
有啥区别

求解�?br />vczh.Iskandar  21:49:27
不是一样吗
御虚舟北(314969051)  21:49:40
改代码中�Q?nbsp;ing
VCZH.�_�丝数组[0]  21:49:49
Windows上的实现是一��L��
May fail spuriously, so must generally be retried in a loop.
�q�一句怎么理解�?br />vczh.Iskandar  21:50:07
compare_exchange_weak
compare_exchange_weak
质量最大vczh�_?402740419)  21:50:14
compare_exchange_weak
compare_exchange_weak

VCZH.�_�丝数组[0]  21:50:16
strong

compare_exchange_strong
�q�有一个问�?br />class spinlock {
private:
  typedef enum {Locked, Unlocked} LockState;
  boost::atomic state_;

public:
  spinlock() : state_(Unlocked) {}

  void lock()
  {
    while (state_.exchange(Locked, boost::memory_order_acquire) == Locked) {
      /* busy-wait */
    }
  }
  void unlock()
  {
    state_.store(Unlocked, boost::memory_order_release);
  }
};

boost例子�l�的 spinloc
怎么是这样实现的
都没有用cae操作
VCZH.�_�丝数组[0]  21:51:20
unlock都直接用store�?br />vczh.Iskandar  21:51:50
不用compare
VCZH.�_�丝数组[0]  21:51:59
为啥
无法理解
vczh.Iskandar  21:52:34
惌��解释好麻�?br />VCZH.�_�丝数组[0]  21:52:40
�q�有在Windows�?br />boost::memory_order_acquire
�q�个参数也没�?br />貌似
求V��解�?br />�q�有strong和weak的区�?br />质量最大vczh�_?402740419)  21:54:46
spinlock本来��׃��用compare�?br />直接swap��p��?br />while (state_.swap(1) == 1);
VCZH.�_�丝数组[0]  21:56:24
你看假设现在是lock状�?br />boost的实现是无条件吧lock换成unlock
如果是��l�lock 他还是lock
VCZH.�_�丝数组[0]  21:58:08
只要要避�?nbsp;重入�?br />lock之前��查一�?br />御虚舟北(314969051)  22:00:03
��康你的书收到没�?br />质量最大vczh�_?402740419)  22:03:17
VCZH.�_�丝数组[0]  21:58:08
只要要避�?nbsp;重入�?br />lock之前��查一�?br />
你用错了他不��你
��是�q�个意�?br />同一个线�E�lock两次也会死锁
VCZH.�_�丝数组[0]  22:05:05
原来是这�?br />
但是他lock的时候不��查，也会��D��两个�U�程同时lock吧？
while (state_.exchange(Locked, boost::memory_order_acquire) == Locked) {
      /* busy-wait */
    }

质量最大vczh�_?402740419)  22:05:18
不会�?br />假设A�q�程先来了，lock成功�?br />VCZH.�_�丝数组[0]  22:05:33
怎能理解�Q�我理解和直接赋值是一�?br />我再�ȝ��
质量最大vczh�_?402740419)  22:05:40
�q�不是复�?br />是exchange
swap
另一个线�E�exchange��׃��收到Locked
那么另一个线�E�就会while循环�Q�直到原来线�E�给讄��了Unlocked
VCZH.�_�丝数组[0]  22:06:47
Exchange current value with new_value, returning current value

exchange是把新值写入旧��D��回么�Q?nbsp; 不是�q�样么？
我有点理解了
质量最大vczh�_?402740419)  22:07:46
对啊�Q�新值写入，旧��D��回，原子�?br />VCZH.�_�丝数组[0]  22:07:59
��是说写入也是写入的lock�Q?nbsp;不媄响之前的lock
当前�U�程拿到旧值检查是不是lock状态，如果是就�l�箋��试直到不是
质量最大vczh�_?402740419)  22:08:00
所以只会有一个线�E�返回Unlocked�Q�另一个线�E�会收到之前�U�程讄��的Locked
VCZH.�_�丝数组[0]  22:08:11
受教�?br />质量最大vczh�_?402740419)  22:08:13
�?br />VCZH.�_�丝数组[0]  22:08:20
我脓到博客上�?nbsp;

Enic 2013-03-31 21:49 发表评论

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boost::atomic 实现 spinlock