c++ - 使用引用作为通用参数从通用 lambda 创建线程

Question

如何使用将自动参数定义为引用的通用 lambda 创建线程？

例如，实现概念上与此等效的东西的正确方法是什么:

int vi = 0;
auto lambda = [](auto &v) {};
auto t = std::thread(lambda, std::ref(vi)); 

gcc-5.3 提示缺少类型:

/opt/gcc/el6/gcc-5.3.0/include/c++/5.3.0/functional: In instantiation of ‘struct std::_Bind_simple<main()::<lambda(auto:2&)>(std::reference_wrapper<int>)>’:
/opt/gcc/el6/gcc-5.3.0/include/c++/5.3.0/thread:137:59:   required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = main()::<lambda(auto:2&)>&; _Args = {std::reference_wrapper<int>}]’
testLambdaCapture.cpp:52:41:   required from here
/opt/gcc/el6/gcc-5.3.0/include/c++/5.3.0/functional:1505:61: error: no type named ‘type’ in ‘class std::result_of<main()::<lambda(auto:2&)>(std::reference_wrapper<int>)>’
   typedef typename result_of<_Callable(_Args...)>::type result_type;
                                                         ^
/opt/gcc/el6/gcc-5.3.0/include/c++/5.3.0/functional:1526:9: error: no type named ‘type’ in ‘class std::result_of<main()::<lambda(auto:2&)>(std::reference_wrapper<int>)>’
         _M_invoke(_Index_tuple<_Indices...>)
         ^

作为附带问题，为什么当泛型参数像这样按值传递时它会起作用:

auto lambda = [](auto v) {};
auto t = std::thread(lambda, vi);

Answer 1

固定:

#include <thread>

int vi = 0;
auto lambda = [](auto &&v) {};
auto t = std::thread(lambda, std::ref(vi)); 

// this works too
auto rv = std::ref(vi);
auto t2 = std::thread(lambda, rv); 

在这种情况下，auto&& 被推导出来，就好像它是一个模板参数一样。因此，实例化期间的实际类型是 const T& 或 T&&(根据需要)。