提高互斥锁顺序 [英] Boost mutex order

问题描述

所以上课简单

class mySafeData
{
public:
  mySafeData() : myData(0)
  {
  }

void Set(int i) 
  {
    boost::mutex::scoped_lock lock(myMutex);
    myData = i; // set the data
    ++stateCounter;  // some int to track state chages
    myCondvar.notify_all(); // notify all readers
  }

  void Get( int& i)
  {
    boost::mutex::scoped_lock lock(myMutex);
    // copy the current state
    int cState = stateCounter;
    // waits for a notification and change of state
    while (stateCounter == cState)
      myCondvar.wait( lock );
  }
 private:
   int myData;
   int stateCounter;
   boost::mutex myMutex;
};

和无限循环中的线程数组调用每个函数

and array of threads in infinite loops calling each one function

 Get()
 Set()
 Get()
 Get()
 Get()

它们将始终以相同的顺序调用函数，并且每个循环仅调用一次(按圆圈，我的意思是所有升压线程每次都将以相同的顺序运行，因此每个线程在一个Set( ))?

will they always call functions in the same order and only once per circle (by circle I mean will all boost threads run in same order each time so that each thread would Get() only once after one Set())?

推荐答案

线程应该以到达scoped_lock构造函数的顺序来获取锁(我认为).但是并不能保证他们会以任何固定顺序到达这一点！

The threads should acquire the lock in the same order that they reach the scoped_lock constructor (I think). But there's no guarantee that they will reach that point in any fixed order!

因此，一般来说:不要依赖它.

So in general: don't rely on it.

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