在C++ 11中通过引用传递对象到std :: thread

为什么在创建时不能通过引用传递对象std::thread？

例如,以下snippit给出了编译错误:

#include 
#include 

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’:
/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’
       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’
         _M_invoke(_Index_tuple<_Indices...>)
         ^

我改变了传递指针,但有更好的解决方法吗？

明确初始化一个线程reference_wrapper使用std::ref:

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

(或视情况std::cref而std::ref定).来自cppreference的std:thread注释:

线程函数的参数按值移动或复制.如果需要将引用参数传递给线程函数,则必须将其包装(例如,使用std::ref或std::cref).

基于此注释，此答案详细说明了默认情况下参数未通过引用传递给线程函数的原因。

考虑以下功能SimpleThread()：

void SimpleThread(int& i) {
    std::this_thread::sleep_for(std::chrono::seconds{1});
    i = 0;
}

现在，想象一下如果编译以下代码（不编译）会发生什么：

int main()
{
    {
        int a;
        std::thread th(SimpleThread, a);
        th.detach();
    }
    // "a" is out of scope
    // at this point the thread may be still running
    // ...
}