c++ - 以成员函数启动线程

Question

我正在尝试构造一个 std::thread，其成员函数不接受任何参数并返回 void。我想不出任何有效的语法——编译器无论如何都会提示。实现 spawn() 以返回执行 test() 的 std::thread 的正确方法是什么？

#include <thread>
class blub {
  void test() {
  }
public:
  std::thread spawn() {
    return { test };
  }
};

Answer 1

#include <thread>
#include <iostream>

class bar {
public:
  void foo() {
    std::cout << "hello from member function" << std::endl;
  }
};

int main()
{
  std::thread t(&bar::foo, bar());
  t.join();
}

编辑:考虑到您的编辑，您必须这样做:

  std::thread spawn() {
    return std::thread(&blub::test, this);
  }

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更新:我想再解释一些要点，其中一些也已在评论中讨论过。

上述语法是根据 INVOKE 定义 (§20.8.2.1) 定义的:

Define INVOKE (f, t1, t2, ..., tN) as follows:

• (t1.*f)(t2, ..., tN) when f is a pointer to a member function of a class T and t1 is an object of type T or a reference to an object of type T or a reference to an object of a type derived from T;
• ((*t1).*f)(t2, ..., tN) when f is a pointer to a member function of a class T and t1 is not one of the types described in the previous item;
• t1.*f when N == 1 and f is a pointer to member data of a class T and t 1 is an object of type T or a
  reference to an object of type T or a reference to an object of a
  type derived from T;
• (*t1).*f when N == 1 and f is a pointer to member data of a class T and t 1 is not one of the types described in the previous item;
• f(t1, t2, ..., tN) in all other cases.

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我想指出的另一个一般事实是，默认情况下，线程构造函数将复制传递给它的所有参数。这样做的原因是参数可能需要比调用线程长，复制参数可以保证这一点。相反，如果您想真正传递一个引用，您可以使用由 std::ref 创建的 std::reference_wrapper。

std::thread (foo, std::ref(arg1));

通过这样做，您 promise 当线程对参数进行操作时，您将确保参数仍然存在。

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请注意，上面提到的所有内容也可以应用于 std::async 和 std::bind。