c++ - 在 C++11 中通过对 std::thread 的引用传递对象

Question

为什么在创建 std::thread 时不能通过引用传递对象？

例如下面的代码片段给出了一个编译错误:

#include <iostream>
#include <thread>

using namespace std;

static void SimpleThread(int& a)  // compile error
//static void SimpleThread(int a)     // OK
{
    cout << __PRETTY_FUNCTION__ << ":" << a << endl;
}

int main()
{
    int a = 6;

    auto thread1 = std::thread(SimpleThread, a);

    thread1.join();
    return 0;
}

错误:

In file included from /usr/include/c++/4.8/thread:39:0,
                 from ./std_thread_refs.cpp:5:
/usr/include/c++/4.8/functional: In instantiation of ‘struct std::_Bind_simple<void (*(int))(int&)>’:
/usr/include/c++/4.8/thread:137:47:   required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(int&); _Args = {int&}]’
./std_thread_refs.cpp:19:47:   required from here
/usr/include/c++/4.8/functional:1697:61: error: no type named ‘type’ in ‘class std::result_of<void (*(int))(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<void (*(int))(int&)>’
         _M_invoke(_Index_tuple<_Indices...>)
         ^

我已更改为传递指针，但是否有更好的解决方法？

Answer 1

使用 reference_wrapper by using std::ref 显式初始化线程:

auto thread1 = std::thread(SimpleThread, std::ref(a));

(或 std::cref 而不是 std::ref，视情况而定)。根据 cppreference on std:thread 的注释:

The arguments to the thread function are moved or copied by value. If a reference argument needs to be passed to the thread function, it has to be wrapped (e.g. with std::ref or std::cref).