通过引用传递数组到线程

Question

我有一个数组需要传递给多个线程进行处理。数组大小在编译时已知。这是直接调用函数而不是通过线程调用的基本要素的工作缩减：

#include <thread>

template <class TYPE>
void function(TYPE& data, std::size_t row, std::size_t column){
    data[row-1][column-1]=2;
}

int main(){
    const int column(5),row(2);
    std::array<std::array<int, column>, row> arr;
    function(arr, row, column);
    return data[row-1][column-1];
}

代码returns2，符合预期。如果我通过

调用函数

    std::thread worker(function<std::array<std::array<int,column>,row>>, arr, row, column);

我收到以下编译器错误：

g++ Testo.cpp -o Testo -std=c++11 -lpthread
In file included from /usr/include/c++/4.8/thread:39:0,
                 from Testo.cpp:2:
/usr/include/c++/4.8/functional: In instantiation of ‘struct std::_Bind_simple<void (*(std::array<std::array<int, 5ul>, 4ul>, int, int))(std::array<std::array<int, 5ul>, 4ul>&, long unsigned int, long unsigned int)>’:
/usr/include/c++/4.8/thread:137:47:   required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(std::array<std::array<int, 5ul>, 4ul>&, long unsigned int, long unsigned int); _Args = {std::array<std::array<int, 5ul>, 4ul>&, const int&, const int&}]’
Testo.cpp:13:82:   required from here
/usr/include/c++/4.8/functional:1697:61: error: no type named ‘type’ in ‘class std::result_of<void (*(std::array<std::array<int, 5ul>, 4ul>, int, int))(std::array<std::array<int, 5ul>, 4ul>&, long unsigned int, long unsigned 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 (*(std::array<std::array<int, 5ul>, 4ul>, int, int))(std::array<std::array<int, 5ul>, 4ul>&, long unsigned int, long unsigned int)>’
         _M_invoke(_Index_tuple<_Indices...>)
         ^

我可以将模板更改为

template <class TYPE>
void function(TYPE data, std::size_t row, std::size_t column){
    data[row-1][column-1]=2;
}

并且编译器不再抱怨，但是 main 然后 returns 0 因为数组不再通过引用传递。线程调用正确模板以通过引用传递数组的适当第一个参数是什么，就像直接调用函数时一样？

Answer 1

通过引用传递 std::ref(arr)

Answer 2

您的函数需要对数组的引用，但 std::thread 存储绑定参数的衰减副本。数组到指针的衰减导致指针类型的纯右值。要修复，请通过 std::reference_wrapper 传递 arr 以保留类型：

std::thread worker(..., std::ref(arr), row, column);

您可能还想使用 lambda，这样就不必显式传递模板参数：

std::thread worker([&arr]{ return function(arr, row, column); });

通过引用传递数组到线程

Passing an array by reference to thread

c++

multithreading

templates

reference

c++11