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线程安全队列死锁

[英]Thread Safe Queue Deadlocks

I have written a thread safe queue which gives deadlock error. 我写了一个线程安全队列,它给出了死锁错误。 I am unable to figure out reason. 我无法找出原因。 I modified the functions to use local locks, instead of member-variable lock. 我修改了函数以使用本地锁,而不是成员变量锁。 Then, it seems to run fine. 然后,它似乎运行良好。

Code: 码:

template <typename T>
class MyQueue {
queue<T> arr;

mutex mtx;
unique_lock<mutex> lck;
condition_variable cv;
public:
MyQueue() {
    lck = unique_lock<mutex>(mtx, defer_lock);
}

void push(int tmp) {
    lck.lock();
    arr.push(tmp);
    lck.unlock();
    cv.notify_one();
}

int pop() {
    T x;
    lck.lock();
    while(arr.size() == 0)
        cv.wait(lck);
    x = arr.front();
    arr.pop();
    lck.unlock();
    return x;
}

int getCount() {
    T x;
    lck.lock();
    x = arr.size();
    lck.unlock();

    return x;
}
};

Error: 错误:

libc++abi.dylib: libc++abi.dylib: libc++abi.dylib: terminating with
uncaught exception of type std::__1::system_error: unique_lock::lock: already locked:
Resource deadlock avoidedterminating with uncaught exception of type std::__1::system_error: 
unique_lock::lock: already locked: Resource deadlock avoidedlibc++abi.dylib: 
terminating with uncaught exception of type std::__1::system_error: unique_lock::lock: already locked: Resource deadlock avoided

As per my comment: unique_lock, as the name implies, is intended for use by only /one/ locking thread. 根据我的评论:顾名思义,unique_lock仅供/ one /锁定线程使用。 To lock from another thread, you need another lock. 要从另一个线程锁定,您需要另一个锁定。 The upshot of this - make unique_lock a local in each function, and not a class member. 这样的结果-在每个函数中使unique_lock成为本地变量,而不是类成员。

template <typename T>
class MyQueue {
queue<T> arr;

mutex mtx;
condition_variable cv;
public:
MyQueue() {
}

void push(int tmp) {
    unique_lock<mutex> lck(mtx);
    arr.push(tmp);
    cv.notify_one();
    lck.unlock(); // Not nec'y, but polite...
}

...

and so forth. 等等。

Consider deleting the unique_lock inside class, and changing your push and pop function to the following: 考虑删除类内部的unique_lock ,并将pushpop函数更改为以下内容:

void push(int tmp)
{
    std::lock_guard<std::mutex> lkg(mtx); // Here unique_lock not necessary.
    arr.push(tmp);
    cv.notify_one();
}

int pop()
{
    std::unique_lock<std::mutex> ulk(mtx);
    cv.wait(ulk, [this]() { return arr.size() != 0; });

    auto x = arr.front();
    arr.pop();
    return x;
}

The reasons are very well explained in the comments :-). 原因在注释中得到了很好的解释:-)。

You might also want to change your mtx to mutable if you want to provide const member functions like empty() ; 如果您想提供const成员函数,例如empty() ,则可能还需要将mtx更改为mutable or copy constructor. 或复制构造函数。

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