锁定自由原子状态类 - 是正确的吗?

时间:2012-04-18 13:55:21

标签: c++ qt locking atomic test-and-set

我只是在尝试在结构上实现原子读/写时寻找反馈(明显的缺陷/改进方法)。

将有一个写线程和多个读线程。目的是防止读者对结构产生不一致的看法,同时不要过多地阻碍作者。

我正在使用fetch-and-add atomic原语,在这种情况下由Qt框架提供。

例如:

/* global */
OneWriterAtomicState<Point> atomicState;

/* Writer */
while(true) {  
  MyStruct s = atomicState.getState()
  s.x += 2; s.y += 2;
  atomicState.setState(s);
}

/* Reader */
while(true) {  
  MyStruct s = atomicState.getState()
  drawBox(s.x,s.y);
}

OneWriterAtomicState实施:

template <class T>
class OneWriterAtomicState
{
public:
    OneWriterAtomicState()
        : seqNumber(0)
    {
    }

    void setState(T& state) {
        this->seqNumber.fetchAndAddOrdered(1);
        this->state = state;
        this->seqNumber.fetchAndAddOrdered(1);
    }

    T getState(){
        T result;
        int seq;
        bool seq_changed = true;

        /* repeat while seq is ODD or if seq changes during read operation */
        while( (seq=this->seqNumber.fetchAndAddOrdered(0)) & 0x01 || seq_changed ) {
            result = this->state;
            seq_changed = (this->seqNumber.fetchAndAddOrdered(0)!=seq);
        }
        return result;
    }


private:
    QAtomicInt seqNumber;
    T state;
} 

这是第二版(memcpy,读者屈服,希望修复getState()):

template <class T>
class OneWriterAtomicState
{

public:
    OneWriterAtomicState()
        : seqNumber(0)
    {
        /* Force a compile-time error if T is NOT a type we can copy with memcpy */
        Q_STATIC_ASSERT(!QTypeInfo<T>::isStatic);
    }

    void setState(T* state) {
        this->seqNumber.fetchAndAddOrdered(1);
        memcpy(&this->state,state,sizeof(T));
        this->seqNumber.fetchAndAddOrdered(1);
    }

    void getState(T* result){
        int seq_before;
        int seq_after  = this->seqNumber.fetchAndAddOrdered(0);
        bool seq_changed = true;
        bool firstIteration = true;

        /* repeat while seq_before is ODD or if seq changes during read operation */
        while( ((seq_before=seq_after) & 0x01) || seq_changed ) {

            /* Dont want to yield on first attempt */
            if(!firstIteration) {
                /* Give the writer a chance to finish */
                QThread::yieldCurrentThread();
            } else firstIteration = false;

            memcpy(result,&this->state,sizeof(T));
            seq_after = this->seqNumber.fetchAndAddOrdered(0);
            seq_changed = (seq_before!=seq_after);
        }
    }

    bool isInitialized() {  return (seqNumber>0); }

private:
    QAtomicInt seqNumber;
    T state;
} ;

#endif // ONEWRITERATOMICSTATE_H

3 个答案:

答案 0 :(得分:2)

算法不太正确。这是一个可能的线程交错,读者获得不一致的数据:

state initialized to {0,0} and seqNumber to 0

Writer:
seqNumber = 1;
state.x = 1;

Reader:
seq = seqNumber; //1
result = state; //{1,0}
seq_changed = (seqNumber != seq); //false

Writer:
state.y = 1;
seqNumber = 2;

Reader:
jumps back to the start of the loop
seq = seqNumber; //2
steps out of the loop because seq == 2 and seq_changed == false

所以问题是在两个地方读取seqNumber并且编写者可以更新读取之间的值。

while( (seq=this->seqNumber.fetchAndAddOrdered(0)) & 0x01 || seq_changed ) {
    result = this->state;
    seq_changed = (this->seqNumber.fetchAndAddOrdered(0)!=seq);
    //If writer updates seqNumber here to even number bad things may happen
}

每次迭代只应读取一次:

T getState(){
    T result;
    int seq;
    int newseq = seqNumber.fetchAndAddOrdered(0);
    bool seq_changed = true;

    while( (seq = newseq) & 0x01 || seq_changed ) {
        result = state;
        newseq = seqNumber.fetchAndAddOrdered(0);
        seq_changed = (newseq != seq);
    }
    return result;
}

我相信这应该正常,但我不保证任何事情。 :)至少你应该编写一个测试程序,就像你的例子中的一个,但在读者中添加一个不一致值的检查。

值得考虑的一件事是使用原子增量(fetchAndAdd)有点过分。只有一个线程写seqNumber,因此您可以使用简单的原子存储释放和负载获取操作,并且可以在许多处理器上更快地实现它们。但是我不知道QAtomicInt是否可以进行这些操作;文件很不清楚。

编辑:和wilx是对的,T需要是一个简单的可复制类型

答案 1 :(得分:1)

我认为只有T的复制赋值运算符是原始的并且基本上只进行按位复制时才会有效。对于任何更复杂的T,您最终可能会在执行result = this->state;期间获得不一致的状态。

所以,我建议使用某种具有作家偏好的rwlock。

答案 2 :(得分:1)

如果您有基于优先级的线程调度,并且阅读器的优先级高于编写器,则可能会遇到活锁。想象一下,作者开始写入价值,然后读者进入主动等待。由于读者的优先级较高,作者永远不会有机会完成写作。

解决方案是在等待循环中添加一个小延迟。