[使用函数对象的C ++线程，如何调用多个析构函数，而不是构造函数？

Question

请在下面找到代码段：-

class tFunc{
    int x;
    public:
    tFunc(){
        cout<<"Constructed : "<<this<<endl;
        x = 1;
    }
    ~tFunc(){
        cout<<"Destroyed : "<<this<<endl;
    }
    void operator()(){
        x += 10;
        cout<<"Thread running at : "<<x<<endl;
    }
    int getX(){ return x; }
};

int main()
{
    tFunc t;
    thread t1(t);
    if(t1.joinable())
    {
        cout<<"Thread is joining..."<<endl;
        t1.join();
    }
    cout<<"x : "<<t.getX()<<endl;
    return 0;
}

我得到的输出是：

Constructed : 0x7ffe27d1b0a4                                                                                                    
Destroyed : 0x7ffe27d1b06c                                                                                                      
Thread is joining...                                                                                                            
Thread running at : 11                                                                                                          
Destroyed : 0x2029c28                                                                                                           
x : 1                                                                                                                               
Destroyed : 0x7ffe27d1b0a4

我很困惑如何调用地址为0x7ffe27d1b06c和0x2029c28的析构函数，而没有调用任何构造函数？而第一个和最后一个构造函数和析构函数分别属于我创建的对象。

Answer 1

您缺少默认的复制构造。对您的程序进行简单的修改，即可提供正在构建的证据。

复制构造函数

#include <iostream>
#include <thread>
#include <functional>
using namespace std;

class tFunc{
    int x;
public:
    tFunc(){
        cout<<"Constructed : "<<this<<endl;
        x = 1;
    }
    tFunc(tFunc const& obj) : x(obj.x)
    {
        cout<<"Copy constructed : "<<this<< " (source=" << &obj << ')' << endl;
    }

    ~tFunc(){
        cout<<"Destroyed : "<<this<<endl;
    }

    void operator()(){
        x += 10;
        cout<<"Thread running at : "<<x<<endl;
    }
    int getX() const { return x; }
};

int main()
{
    tFunc t;
    thread t1{t};
    if(t1.joinable())
    {
        cout<<"Thread is joining..."<<endl;
        t1.join();
    }
    cout<<"x : "<<t.getX()<<endl;
    return 0;
}

输出（地址不同）

Constructed : 0x104055020
Copy constructed : 0x104055160 (source=0x104055020)
Copy constructed : 0x602000008a38 (source=0x104055160)
Destroyed : 0x104055160
Thread running at : 11
Destroyed : 0x602000008a38
Thread is joining...
x : 1
Destroyed : 0x104055020

---

复制构造函数和移动构造函数

如果您提供搬家公司，则至少应使用其中一份复印件：

#include <iostream>
#include <thread>
#include <functional>
using namespace std;

class tFunc{
    int x;
public:
    tFunc(){
        cout<<"Constructed : "<<this<<endl;
        x = 1;
    }
    tFunc(tFunc const& obj) : x(obj.x)
    {
        cout<<"Copy constructed : "<<this<< " (source=" << &obj << ')' << endl;
    }

    tFunc(tFunc&& obj) : x(obj.x)
    {
        cout<<"Move constructed : "<<this<< " (source=" << &obj << ')' << endl;
        obj.x = 0;
    }

    ~tFunc(){
        cout<<"Destroyed : "<<this<<endl;
    }

    void operator()(){
        x += 10;
        cout<<"Thread running at : "<<x<<endl;
    }
    int getX() const { return x; }
};

int main()
{
    tFunc t;
    thread t1{t};
    if(t1.joinable())
    {
        cout<<"Thread is joining..."<<endl;
        t1.join();
    }
    cout<<"x : "<<t.getX()<<endl;
    return 0;
}

输出（地址不同）

Constructed : 0x104057020
Copy constructed : 0x104057160 (source=0x104057020)
Move constructed : 0x602000008a38 (source=0x104057160)
Destroyed : 0x104057160
Thread running at : 11
Destroyed : 0x602000008a38
Thread is joining...
x : 1
Destroyed : 0x104057020

---

参考包装

最后，如果要避免这些副本，可以将可调用对象包装在参考包装中（std::ref）。由于您想在穿线部分完成后再使用t，因此这对您来说是可行的。实际上，在对调用对象的引用进行线程化时，必须非常小心，因为对象的生存期至少必须延长，只要线程利用引用即可。

#include <iostream>
#include <thread>
#include <functional>
using namespace std;

class tFunc{
    int x;
public:
    tFunc(){
        cout<<"Constructed : "<<this<<endl;
        x = 1;
    }
    tFunc(tFunc const& obj) : x(obj.x)
    {
        cout<<"Copy constructed : "<<this<< " (source=" << &obj << ')' << endl;
    }

    tFunc(tFunc&& obj) : x(obj.x)
    {
        cout<<"Move constructed : "<<this<< " (source=" << &obj << ')' << endl;
        obj.x = 0;
    }

    ~tFunc(){
        cout<<"Destroyed : "<<this<<endl;
    }

    void operator()(){
        x += 10;
        cout<<"Thread running at : "<<x<<endl;
    }
    int getX() const { return x; }
};

int main()
{
    tFunc t;
    thread t1{std::ref(t)}; // LOOK HERE
    if(t1.joinable())
    {
        cout<<"Thread is joining..."<<endl;
        t1.join();
    }
    cout<<"x : "<<t.getX()<<endl;
    return 0;
}

输出（地址不同）

Constructed : 0x104057020
Thread is joining...
Thread running at : 11
x : 11
Destroyed : 0x104057020

请注意，尽管我保留了copy-ctor和move-ctor重载，但我们都没有调用它们，因为引用包装器现在是要复制/移动的东西；不是它引用的东西。同样，这种最终方法可以满足您的需求。实际上，将t.x中的main修改为11。以前没有尝试过。压力不够。小心这样做。对象生存期为critical。