Your first line printme({‘a’, ‘b’, ‘c’}) is illegal because the template argument T could not be inferred. If you explicitly specify the template argument it will work, e.g. printme<vector<char>>({‘a’, ‘b’, ‘c’}) or printme<initializer_list<char>>({‘a’, ‘b’, ‘c’}). The other ones you listed are legal because the argument has a well-defined type, so the template argument T can … Read more
I am assuming that you are using a C99 compiler (with support for dynamically sized arrays). The problem in your code is that at the time when the compilers sees your variable declaration it cannot know how many elements there are in the array (I am also assuming here, from the compiler error that length … Read more
Indeed the final version of C++11 does not enable the use of initializer lists on the right-hand side (or left-hand side, for that matter) of a binary operator. Firstly, initializer-lists are not expressions as defined in §5 of the Standard. The arguments of functions, as well as of binary operators, generally have to be expressions, … Read more
No, that won’t work as intended; you will still get copies. I’m pretty surprised by this, as I’d thought that initializer_list existed to keep an array of temporaries until they were move‘d. begin and end for initializer_list return const T *, so the result of move in your code is T const && — an … Read more
Edit: Since @Johannes doesn’t seem to want to post the best solution as an answer, I’ll just do it. #include <iterator> #include <vector> #include <memory> int main(){ using move_only = std::unique_ptr<int>; move_only init[] = { move_only(), move_only(), move_only() }; std::vector<move_only> v{std::make_move_iterator(std::begin(init)), std::make_move_iterator(std::end(init))}; } The iterators returned by std::make_move_iterator will move the pointed-to element when being … Read more