问题
Possible Duplicate:
initializer_list and move semantics
In this code:
#include <vector>
#include <initializer_list>
template<typename T>
class some_custom_container : public std::vector<T>
{
public:
some_custom_container(const std::initializer_list<T>& contents)
{
for (auto& i : contents)
this->emplace_back(std::move(i));
}
};
class test_class
{};
int main()
{
test_class a;
some_custom_container<test_class> i = { a, test_class(), a };
}
If I've understand it, all objects in { a, test_class(), a } are safe-constructed: the named-objects are copied and the unnamed objects are moved to construct the initializer_list. After, this initializer_list is passed by reference to the some_custom_container's constructor.
Then, to avoid useless doble-copies I move all of them to fill the vector.
Is it safe this constructor? I mean, in a strange situation, for example if T is evaluated as a reference & or &&, is the vector always well-filled (contains it safe objects)?
If this is the case, why the initializer_list constructor implementations of stl containers aren't implemented in this way? As I know, their constructors copy and don't move the contents.
回答1:
initializer_list only provides const access to its elements. You could use const_cast to make that code compile, but then the moves might end up with undefined behaviour (if the elements of the initializer_list are truly const). So, no it is not safe to do this moving. There are workarounds for this, if you truly need it.
来源：https://stackoverflow.com/questions/13957166/is-it-safe-to-move-elements-of-a-initializer-list