请找到下面的代码片段:
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构造函数?而第一个和最后一个构造函数和析构函数分别是我创建的对象的。
最佳答案
您缺少检测复制构造和移动构造。对程序进行简单修改即可提供构造发生位置的证据。
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
<小时/>
复制构造函数和移动构造函数
如果您提供一个移动 vector ,那么至少其中一个拷贝将是首选:
#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
<小时/>
引用包装
如果您想避免这些拷贝,您可以将可调用对象包装在引用包装器 (std::ref
) 中。由于您希望在线程部分完成后使用 t
,因此这对于您的情况是可行的。实际上,在针对调用对象的引用进行线程处理时,您必须非常小心,因为对象的生命周期必须至少与使用该引用的线程一样长。
#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,尽管我保留了复制 vector 和移动 vector 重载,但两者都没有被调用,因为引用包装器现在是被复制/移动的东西;不是它引用的东西。此外,最后的方法提供了您可能正在寻找的内容;实际上,main
中的 t.x
已修改为 11
。这在之前的尝试中是没有的。然而,这一点怎么强调都不为过:这样做要小心。对象生命周期至关重要。
移动,无非
最后,如果您没有兴趣像示例中那样保留 t
,则可以使用移动语义将实例直接发送到线程,并一路移动。
#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()
{
thread t1{tFunc()}; // LOOK HERE
if(t1.joinable())
{
cout<<"Thread is joining..."<<endl;
t1.join();
}
return 0;
}
输出(地址不同)
Constructed : 0x104055040
Move constructed : 0x104055160 (source=0x104055040)
Move constructed : 0x602000008a38 (source=0x104055160)
Destroyed : 0x104055160
Destroyed : 0x104055040
Thread is joining...
Thread running at : 11
Destroyed : 0x602000008a38
在这里你可以看到对象被创建,对 said-same 的右值引用然后直接发送到 std::thread::thread()
,在那里它再次移动到其最终的休息位置,从该点开始由线程拥有。不涉及抄袭者。实际的 dtors 针对两个外壳和最终目标具体对象。
关于C++线程使用函数对象,如何调用多个析构函数而不调用构造函数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59064337/