Why is std::queue not thread-safe? - c++

Question

The topic says it. I don't understand why the std::queue (or in general: any queue) is not thread-safe by its nature, when there is no iterator involved as with other datastructures.
According to the common rule that
at least one thread is writing to ...
and another thread is reading from a shared resource
I should have gotten a conflict in the following example code:
#include "stdafx.h"
#include <queue>
#include <thread>
#include <iostream>
struct response
{
static int & getCount()
{
static int theCount = 0;
return theCount;
}
int id;
};
std::queue<response> queue;
// generate 100 response objects and push them into the queue
void produce()
{
for (int i = 0; i < 100; i++)
{
response r;
r.id = response::getCount()++;
queue.push(r);
std::cout << "produced: " << r.id << std::endl;
}
}
// get the 100 first responses from the queue
void consume()
{
int consumedCounter = 0;
for (;;)
{
if (!queue.empty())
{
std::cout << "consumed: " << queue.front().id << std::endl;
queue.pop();
consumedCounter++;
}
if (consumedCounter == 100)
break;
}
}
int _tmain(int argc, _TCHAR* argv[])
{
std::thread t1(produce);
std::thread t2(consume);
t1.join();
t2.join();
return 0;
}
Everything seems to be working fine:
- No integrity violated / data corrupted
- The order of the elements in which the consumer gets them are correct (0<1<2<3<4...), of course the order in which the prod. and cons. are printing is random as there is no signaling involved.

Answer 1

Imagine you check for !queue.empty(), enter the next block and before getting to access queue.first(), another thread would remove (pop) the one and only element, so you query an empty queue.
Using a synchronized queue like the following
#pragma once
#include <queue>
#include <mutex>
#include <condition_variable>
template <typename T>
class SharedQueue
{
public:
SharedQueue();
~SharedQueue();
T& front();
void pop_front();
void push_back(const T& item);
void push_back(T&& item);
int size();
bool empty();
private:
std::deque<T> queue_;
std::mutex mutex_;
std::condition_variable cond_;
};
template <typename T>
SharedQueue<T>::SharedQueue(){}
template <typename T>
SharedQueue<T>::~SharedQueue(){}
template <typename T>
T& SharedQueue<T>::front()
{
std::unique_lock<std::mutex> mlock(mutex_);
while (queue_.empty())
{
cond_.wait(mlock);
}
return queue_.front();
}
template <typename T>
void SharedQueue<T>::pop_front()
{
std::unique_lock<std::mutex> mlock(mutex_);
while (queue_.empty())
{
cond_.wait(mlock);
}
queue_.pop_front();
}
template <typename T>
void SharedQueue<T>::push_back(const T& item)
{
std::unique_lock<std::mutex> mlock(mutex_);
queue_.push_back(item);
mlock.unlock(); // unlock before notificiation to minimize mutex con
cond_.notify_one(); // notify one waiting thread
}
template <typename T>
void SharedQueue<T>::push_back(T&& item)
{
std::unique_lock<std::mutex> mlock(mutex_);
queue_.push_back(std::move(item));
mlock.unlock(); // unlock before notificiation to minimize mutex con
cond_.notify_one(); // notify one waiting thread
}
template <typename T>
int SharedQueue<T>::size()
{
std::unique_lock<std::mutex> mlock(mutex_);
int size = queue_.size();
mlock.unlock();
return size;
}
The call to front() waits until it has an element and locks the underlying queue so only one thread may access it at a time.