Multithreading 为什么是C++；线程多次调用析构函数？ 我很好奇，为什么一个简单的C++程序，像下面这样的一个程序，在处理STD:：线程时，多次调用Work1对象的析构函数。为什么它多次调用析构函数 class work1 { public: work1(int x_) : x(x_) { ++count; std::cout<<"\n[WORKER1] Ctor : "<<count<<std::endl; } ~work1() { std::cout<<"\n[WORKER1] Dtor : "<<count<<std::endl; } void operator() () { for(int i =0; i < x; ++i) { std::cout<<"[WORKER1] Printing this :: "<<i<<std::endl; } } private: int x; static int count; }; int main() { int local_main=5; std::thread t((work1(local_main))); t.join(); std::cout<<"\n[MAIN] From main thread..."<<std::endl; return 0; }

std:：thread

通过调用复制构造函数来复制参数。实施副本构造函数以观察副本：

class work1 {
public:
    work1(int x_) : x(x_) {
        ++count;
        std::cout << __PRETTY_FUNCTION__ << ' ' << count << '\n';
    }

    work1(work1 const& other)
        : x(other.x)
    {
        ++count;
        std::cout << __PRETTY_FUNCTION__ << ' ' << count << '\n';
    }

    work1& operator=(work1 const& other)
    {
        x = other.x;
        std::cout << __PRETTY_FUNCTION__ << ' ' << count << '\n';
        return *this;
    }

    ~work1() {
        --count;
        std::cout << __PRETTY_FUNCTION__ << ' ' << count << '\n';
    }

    void operator() () {
        for(int i =0; i < x; ++i) {
            std::cout<<"[WORKER1] Printing this :: "<<i<<std::endl;
        }
    }

private:
    int x;
    static int count;
};

int work1::count;

int main() {
    int local_main=5;
    std::thread t((work1(local_main)));
    t.join();
}

如果要避免复制，请使用

std:：ref

将引用传递给函数对象：

int main() {
    work1 w{5};
    std::thread t(std::ref(w));
    t.join();
}

产出：

work1::work1(int) 1
work1::work1(const work1&) 2
work1::work1(const work1&) 3
work1::~work1() 2
work1::~work1() 1
[WORKER1] Printing this :: 0
[WORKER1] Printing this :: 1
[WORKER1] Printing this :: 2
[WORKER1] Printing this :: 3
[WORKER1] Printing this :: 4
work1::~work1() 0

work1::work1(int) 1
[WORKER1] Printing this :: 0
[WORKER1] Printing this :: 1
[WORKER1] Printing this :: 2
[WORKER1] Printing this :: 3
[WORKER1] Printing this :: 4
work1::~work1() 0

---

由于我们没有解决方案，我们无法真正告诉您发生了什么，也无法帮助您找出原因以及如何解决任何问题。另外，请花一些时间刷新，以及。它现在更清晰，这只是一个复制（或移动）的情况。线程需要调用函数的副本，在您的情况下，每个副本都是对象的副本（这导致复制或移动构造）然后是可能的临时对象的破坏。我们如何控制并纠正这个问题。但是你不认为抛出同一对象的引用不是一个好主意，因为其他线程也会持有对同一对象的引用。这不是真的吗？很抱歉，我需要澄清这几个概念。@AnubhavRohatgi线程拥有自己的副本是最安全的。对于引用，当另一个线程使用该对象时，该对象必须保持活动状态。

work1::work1(int) 1
[WORKER1] Printing this :: 0
[WORKER1] Printing this :: 1
[WORKER1] Printing this :: 2
[WORKER1] Printing this :: 3
[WORKER1] Printing this :: 4
work1::~work1() 0