I've written a convenient functor wrapper for tuple std::get. When using it with boost transformed and operator[], I get warning that I'm returning reference to local temporary object. My system: ubuntu 14.04, compilers: clang-3.5 and g++-4.8.2, boost version: 1.56.
#include <boost/range/adaptor/transformed.hpp>
#include <utility>
#include <vector>
template <std::size_t I>
struct tuple_get {
template <typename Tuple>
auto operator()(Tuple &&tuple) const ->
decltype(std::get<I>(std::forward<Tuple>(tuple))) {
return std::get<I>(std::forward<Tuple>(tuple));
}
};
int main() {
//removing const gets rid of warning
const std::vector<std::tuple<int,int>> v = {std::make_tuple(0, 0)};
//gives warning
(v | boost::adaptors::transformed(tuple_get<0>{})) [0];
}
Warning details:
include/boost/range/iterator_range_core.hpp:390:16: warning: returning reference to local temporary object [-Wreturn-stack-address]
return this->m_Begin[at];
note: in instantiation of member function 'boost::iterator_range_detail::iterator_range_base<boost::transform_iterator<tuple_get<0>,
std::__1::__wrap_iter<const std::__1::tuple<int, int> *>, boost::use_default, boost::use_default>, boost::random_access_traversal_tag>::operator[]' requested here
(v | boost::adaptors::transformed(tuple_get<0>{})) [0];
Adding flag -Wreturn-stack-address is not a solution since it's dangerous in bigger projects.
I noticed that deleting const keyword gets rid of warning but I don't know why and don't want to assume that functor gets only non-const ranges.
Questions: how to fix code to get rid of warning? Why deleting const gets rid of warning?
It's true.
So, you should be able to fix it with
which returns the
abstract_value_type
(which is the reference only if the elements are abstract, array or function, the value_type otherwise).