c++ - 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。这是对直接调用函数而不是通过线程调用函数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];
}

如预期的那样，代码返回 2。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...>)
         ^

我可以将我的模板更改为

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

并且编译器不再报错，但 main 然后返回 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

您的函数需要对数组的引用，但 std::thread 存储绑定(bind)参数的衰减拷贝。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，这样您就不必显式传递模板参数:

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