C++11中的线程移动

Question

在使用C++并发测试示例时，我遇到了一些问题。

/***
Scoped_thread
Help explain the move semantics of Scoped_guard
@c++ Concurrency in Action
***/
#include <thread>
#include <iostream>
using namespace std;
class Scoped_thread
{
  std::thread t;
public:
  Scoped_thread(std::thread _t):
    t(std::move(_t))
    {
      cout << "Success?" << endl;
      if (!t.joinable())
        throw std::logic_error("No thread");
    }
  ~Scoped_thread()
  {
    t.join();
  }
  Scoped_thread(Scoped_thread const&) = delete;
  Scoped_thread& operator=(Scoped_thread const&) = delete;
};

struct func
{
  int& i;
  func(int& i):i(i) {}
  void operator()()
  {
    for (unsigned j = 0; j < 1000000; j++)
    {
      cout << j << endl;
    }
  }
};

int main()
{
  int some_local_state = 1;
  func myfunc(some_local_state);
  Scoped_thread t2(std::thread(myfunc));
  for (unsigned j = 0; j < 1000; j++)
  {
    cout << "Main thread " << j << endl;
  }
}

打印输出时，仅显示“主线程”。我发现构造函数没有启动。这是否表明使用线程移动语义存在问题？我的工作环境是Ubuntu16.04，编译命令是'g++ -std=c++11 -Wall -pthread file.cpp‘

Answer 1

Scoped_thread t2(std::thread(myfunc));

这里我们有一个稍微非常规的most vexing parse案例。问题是:下面的函数转发声明是等价的：

void f(int arg);
void f(int (arg));

因此，Scoped_thread t2(std::thread(myfunc));被解析为函数t2的正向声明，该函数返回Scoped_thread并将std::thread myfunc作为参数。

两种解决方案是：

Scoped_thread t2{std::thread(myfunc)};