将成员函数用于带有 std::ref(*this) 的 std::thread 无法编译

Question

我使用以下最小示例来重现当我尝试创建一个调用非静态成员函数来完成其工作的线程时遇到的编译器错误：

#include <thread>
#include <iostream>
class Worker
{
public:
  Worker() : m_worker(&Worker::doWork, std::ref(*this), 1)
  {}
  std::thread m_worker;

  void doWork(int a) { std::cout << a << std::endl; }
};

int main(int argc, char* argv[]) {
  Worker k;
}

使用 gcc4.8-gcc5.1 时编译失败，原因如下：

In file included from /usr/include/c++/4.8/thread:39:0,

from /tmp/gcc-explorer-compiler115614-69-rgangs/example.cpp:1:

/usr/include/c++/4.8/functional: In instantiation of 'struct std::_Bind_simple<std::_Mem_fn<void (Worker::*)(int)>(std::reference_wrapper<Worker>, int)>':

/usr/include/c++/4.8/thread:137:47: required from 'std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (Worker::*)(int); _Args = {std::reference_wrapper<Worker>, int}]'

7 : required from here

/usr/include/c++/4.8/functional:1697:61: error: no type named 'type' in 'class std::result_of<std::_Mem_fn<void (Worker::*)(int)>(std::reference_wrapper<Worker>, int)>'

typedef typename result_of<_Callable(_Args...)>::type result_type;

^

/usr/include/c++/4.8/functional:1727:9: error: no type named 'type' in 'class std::result_of<std::_Mem_fn<void (Worker::*)(int)>(std::reference_wrapper<Worker>, int)>'

_M_invoke(_Index_tuple<_Indices...>)

^

Compilation failed

另一方面，Clang 似乎可以很好地编译这段代码。谁在这里是正确的，这是 gcc 中的错误吗（有公开的票证？）？

编辑： 当使用 m_worker(&Worker::doWork, this, 1) 初始化线程时，gcc 编译它就好了。那么，在这种情况下使用 std::ref(*this) 合法吗？我想 std::ref() 更笼统。

Answer 1

您的 thread 构造函数调用依赖于以下语义：

[C++14: 30.3.1.2/3]: Requires: F and each Ti in Args shall satisfy the MoveConstructible requirements. INVOKE (DECAY_COPY ( std::forward<F>(f)), DECAY_COPY (std::forward<Args>(args))...) (20.9.2) shall be a valid expression.

而INVOKE是这样定义的：

[C++14: 20.9.2/1]: 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 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 when N == 1 and f is a pointer to member data of a class T and t1 is not one of the types described in the previous item;

f(t1, t2, ..., tN) in all other cases.

如你所见，std::reference_wrapper<Worker>这里没有规定，这是std::ref(*this)给你的。当然，衰减规则中的任何内容都没有帮助 ([C++14: 30.2.6/1])。

Clang 实际上有点仓促行事，似乎允许这样做是因为有一天它会符合标准，这要归功于我们自己的 Jonathan Wakely filing Library Working Group issue #2219。但是，目前还没有。

无论如何，这整件事都没有实际意义。无需编写此代码。就这么写：

Worker() : m_worker(&Worker::doWork, this, 1)

将成员函数用于带有 std::ref(*this) 的 std::thread 无法编译

Use member function for `std::thread` with `std::ref(*this)` fails to compile

c++

gcc

standards-compliance

c++14