我想在某个类的成员变量中保存一个线程。以下代码片段显示了我想要实现的目标:
#include <iostream>
#include <thread>
#include <vector>
class Test {
public:
std::thread& t;
Test(std::thread&& rt) : t(rt) {}
};
int main()
{
std::vector<Test> tests;
{
std::thread t ([]{
std::cout << 1;
});
tests.push_back(Test(std::move(t)));
}
for(Test mytest : tests)
{
mytest.t.join();
}
}
代码将在join()行中断。错误是:
terminate called without an active exception
Aborted (core dumped)
当原始线程创建的范围遗留下来时,为什么我不能通过mytest.t调用该线程?
由于std :: thread是可移动但不可复制的,你可以这样做:
class Test {
public:
std::thread t;
Test(std::thread&& rt) : t(std::move(rt)) {}
};
int main()
{
std::vector<Test> tests;
{
std::thread t([] {
std::cout << 1;
});
tests.push_back(Test(std::move(t)));
}
for (Test& mytest : tests)
{
mytest.t.join();
}
}
在您的类中,您有一个线程的引用,而不是一个线程对象:
std::thread& t;
^
这意味着将发生以下顺序:
{
std::thread t ([]{
std::cout << 1;
}); // 1. Thread is created.
tests.push_back(Test(std::move(t))); // 2. Reference to moved thread is taken
// and after move thread is destroyed.
// 3. Now the thread is destroyed,
// but not joined which will call `std::terminate`
// (Thanks @tkausl)
}
如果你让你的班级std::thread t
移动将有效。
正如@tkausl所提到的,它是一个引用,一旦{}
超出范围并且你的引用不再有效,它就会销毁它。此外,您需要修改循环,以便它不会创建原始Test
对象的副本。修改后,这将成为:
class Test {
public:
std::thread& t;
Test(std::thread&& rt) : t(rt) {}
};
int main()
{
std::vector<Test> tests;
std::thread t ([]{
std::cout << 1;
});
tests.push_back(Test(std::move(t)));
for(Test& mytest : tests)
{
mytest.t.join();
}
}