我學習我的方式,通過與C++ 11的同步原語我必須寫一個模板類,這是那些方法在施工時聲明最大元素數的FIFO隊列。在一個簡單的C++ 11四線程程序時,我的兩個消費者線程沒有返回,如果我的評論標準輸出打印線
有兩個線程這推動項目到所說的隊列中,另兩種檢索。它們通過使用兩個條件變量進行同步,以確保消費者線程僅在隊列不爲空時彈出項目,並且生產者線程僅在隊列未滿時才推送新項目。該隊列已獲得打開/關閉狀態,該狀態在條件變量的wait()調用中用作附加條件。當隊列關閉時,線程應該返回而不執行任何操作。
// main.cpp
#include "stdafx.h"
int _tmain(int argc, _TCHAR* argv[]){
BlockingQueue<int> bq(10);
int sum1=0, sum2=0;
std::thread c1([&bq,&sum1](){
int i;
while(bq.get(i)) sum1+=i;
});
std::thread c2([&bq,&sum2](){
int i;
while(bq.get(i)) sum2+=i;
});
std::thread p1([&bq](){
for(int i=0;i<1000;i+=2) bq.put(i);
});
std::thread p2([&bq](){
for(int i=0;i<1000;i+=2) bq.put(i+1);
});
p1.join();
std::cout<<"p1 thread returned."<<std::endl;
p2.join();
std::cout<<"p2 thread returned."<<std::endl;
bq.close();
c1.join();
std::cout<<"c1 thread returned."<<std::endl;
c2.join();
std::cout<<"c2 thread returned."<<std::endl;
std::cout<<"sum1: "<<sum1<<std::endl;
std::cout<<"sum2: "<<sum2<<std::endl;
std::cout<<"total: "<<sum1+sum2<<std::endl;
return 0;
}
下面是我創建的類:
// BlockingQueue.h
#include "stdafx.h"
template<class T> class BlockingQueue
{
std::mutex t_queue_mutex;
std::queue<T> t_queue;
int t_queue_cap_value;
bool isQueueOpen;
std::condition_variable put_condition;
std::condition_variable get_condition;
public:
BlockingQueue(int N);
~BlockingQueue(void);
bool put(T t_item);
bool get(T &t_item);
bool isOpen();
bool isFull();
bool isEmpty();
void close();
};
// BlockinQueue.cpp
#include "BlockingQueue.h"
#include "stdafx.h"
template <class T> BlockingQueue<T>::BlockingQueue(int N)
{
t_queue_cap_value=N;
isQueueOpen=true;
std::cout<<"Rejoice! A bq has been created!"<<std::endl;
}
template <class T> BlockingQueue<T>::~BlockingQueue(void)
{
}
template <class T> bool BlockingQueue<T>::isFull(){
if(t_queue_cap_value==t_queue.size())
return true;
else
return false;
}
template <class T> bool BlockingQueue<T>::isOpen(){
return isQueueOpen;
}
template <class T> void BlockingQueue<T>::close(){
isQueueOpen=false;
}
/* get method */
template <class T> bool BlockingQueue<T>::get(T &t_item){
bool exitThreadStatus=false;
if(!isOpen()){
put_condition.notify_all();
return false;
}
std::unique_lock<std::mutex> ul(t_queue_mutex);
get_condition.wait(ul, [this](){
//std::cout<<"Getter thread with get_id()="<<std::this_thread::get_id()<<" is waiting. isOpen()="<<isOpen()<<" and t_queue.empty()="<<t_queue.empty()<<std::endl;
if(!isOpen())
return true;
else
return !t_queue.empty();
});
if(isOpen()){
exitThreadStatus=true;
t_item=t_queue.front();
t_queue.pop();
}
std::cout<<"Extracted "<<t_item<<". After pop size()="<<t_queue.size()<<std::endl;
put_condition.notify_all();
return exitThreadStatus;
}
/* put method */
template <class T> bool BlockingQueue<T>::put(T t_item){
bool exitThreadStatus=false;
if(!isOpen()){
get_condition.notify_all();
return false;
}
std::unique_lock<std::mutex> ul(t_queue_mutex);
put_condition.wait(ul, [this](){
if(!isOpen())
return true;
else
return !isFull();
});
if(isOpen()){
exitThreadStatus=true;
t_queue.push(t_item);
}
std::cout<<"Inserting "<<t_item<<". After push size()="<<t_queue.size()<<std::endl;
get_condition.notify_all();
return exitThreadStatus;
}
template class BlockingQueue<int>;
這似乎是正常工作,每當我離開的get(兩個性病::法院線)和put()註釋掉,得到下面的輸出(如預期):
Inserting 998. After push size()=2
Extracted 997. After pop size()=1
p1 thread returned.
Inserting 999. After push size()=2
Extracted 998. After pop size()=1
p2 thread returned.
Extracted 999. After pop size()=0
Extracted 998. After pop size()=0
c1 thread returned.
c2 thread returned.
sum1: 250000
sum2: 249500
total: 499500
如果我不是評論的COUT線,兩線收集似乎永遠回來,我不明白什麼是錯我的代碼。有人有線索嗎?謝謝!
輸出帶有註釋COUT行:
Rejoice! A bq has been created!
p1 thread returned.
p2 thread returned.
這一工程!非常感謝。通過使用喚醒條件作爲傳遞給條件變量wake()調用的lambda表達式形式的第二個參數來處理虛假喚醒。 – wallen 2013-04-21 17:04:17