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。 这里是一个简化的示例,其中函数直接被调用,而不是通过线程:
#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];
}
代码按预期返回2。 如果我通过调用函数
std::thread worker(function<std::array<std::array<int,column>,row>>, arr, row, column);
我收到了以下编译器错误信息:
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...>)
^
我可以更改我的模板为:
template <class TYPE>
void function(TYPE data, std::size_t row, std::size_t column){
data[row-1][column-1]=2;
}
现在编译器已不再报错,但由于数组不再通过引用传递,因此主函数返回0。 那么线程的正确模板首个参数是什么,以便像直接调用函数一样通过引用传递数组?
arr
在你使用它之后才被声明在function
中,那么它应该如何工作呢? - crashmstrarr
在函数调用之后定义。 - NathanOliver