c++ - 这是在 C++ 中实现有界缓冲区的正确方法吗

标签 c++ multithreading vector thread-safety buffer

<分区>

我正在开发一个处理多个线程访问、存入和退出有界缓冲区容器的程序。我注意到线程存在一些主要问题,并怀疑我的缓冲区在某处部分或根本不正确。

为了确保我知道我在用它做什么,我希望能检查一下我的缓冲区代码。该类使用我在其他地方实现的信号量,我假设它现在可以工作(如果不能,我会尽快弄清楚!)我添加了试图解释我的推理的评论。

首先是.h文件:

#ifndef BOUNDED_BUFFER_H
#define BOUNDED_BUFFER_H

#include "Semaphore.H"
#include <string> 
#include <vector>  

using namespace std; 

class Item{ //supposed to eventually be more extensible...seems silly with just a string for now

  public:
    Item(string _content){content = _content;} 
    string GetContent() {return content;}     

  private:  
};   

    class BoundedBuffer{

      public:
        BoundedBuffer(); 

        void Deposit(Item* _item); 
        Item* Retrieve();        
        int GetNumItems() {return count;} 
        vector<Item*> GetBuffer() {return buffer;} 
        void SetSize(int _size){
          capacity = _size;
          buffer.reserve(_size);  //or do I need to call "resize" 
        }  

      private:
        int capacity; 
        vector<Item*> buffer; //I originally wanted to use an array but had trouble with  
                              //initilization/sizing, etc. 
        int nextin; 
            int nextout; 
            int count; 

            Semaphore notfull;   //wait on this one before depositing an item
            Semaphore notempty;  //wait on this one before retrieving an item
        };

    #endif

接下来,.cpp:

#include "BoundedBuffer.H"
#include <iostream>

using namespace std; 

BoundedBuffer::BoundedBuffer(){

  notfull.SetValue(0); 
  notempty.SetValue(0); 
  nextin = 0; 
  nextout = 0; 
  count = 0; 
}

void BoundedBuffer::Deposit(Item* _item){
  if(count == capacity){ 
    notfull.P(); //Cannot deposit into full buffer; wait
  }

  buffer[nextin] = _item; 
  nextin = (nextin + 1) % capacity;  //wrap-around
  count += 1;
  notempty.V();  //Signal that retrieval is safe 
}

Item* BoundedBuffer::Retrieve(){
  if(count == 0){
    notempty.P(); //cannot take from empty buffer; wait 
  }

  Item* x = buffer[nextout]; 
  nextout = (nextout + 1) % capacity;
  buffer.pop_back();  //or a different erase methodology? 
  count -= 1; 
  notfull.V(); //Signal that deposit is safe 
  return x; 
}

我认为问题可能源于我选择 vector 作为底层容器(或者更确切地说,不正确地使用它),或者可能需要更多的安全阻塞机制(互斥锁等?)来自事情的样子,任何人都可以提供一些反馈吗?

最佳答案

这是一个非常常见的问题(关于如何做一个正确的多线程队列)。我之前看到的最好的答案是this stack overflow questionthis web site .这些答案适用于无界队列,因此我将在此处扩展并显示有界队列的答案。

您需要使用互斥锁保护您的存款和检索功能,并使用条件变量来进行唤醒。

#include <mutex>
#include <condition_variable>

std::mutex the_mutex;
std::condition_variable the_notfull_cvar;
std::condition_variable the_notempty_cvar;

...

void BoundedBuffer::Deposit(Item* _item){
  std::unique_lock<std::mutex> lock(the_mutex);
  while ( /* buffer is full */ ){
    /* simultaneously wait and release the mutex */
    the_notfull_cvar.wait(lock);
    /* the mutex is reaquired at this point */
  }

  /* buffer has space and we own the mutex: insert the item */
  ...
  /* tell anyone waiting on an empty buffer that they can wake up. */
  the_notempty_cvar.notify_all();
}

Item* BoundedBuffer::Retrieve(){
  std::unique_lock<std::mutex> lock(the_mutex);
  while ( /* buffer is empty */ ){
    /* simultaneously wait and release the mutex */
    the_notempty_cvar.wait(lock);
    /* the mutex is reaquired at this point */
  }

  /* buffer has something in it and we own the mutex: get the item */
  ...
  /* tell anyone waiting on a full buffer that they can wake up. */
  the_notfull_cvar.notify_all();

  return x;
}

您的 GetNumItems()、GetBuffer() 和 SetSize() 函数也需要使用 unique_locks 进行保护。

关于c++ - 这是在 C++ 中实现有界缓冲区的正确方法吗,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15798222/

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