This is not copy-initializing, or is it?
c++, c++11, copy-initialization, explicit-constructor
Solution
The `=` notation should not affect the complaint because reference binding doesn't behave differently whether expressed by direct- or copy-initialization. What's being initialized here is the return value object, which does not have its own name.
Unfortunately, GCC is right to complain, as does Clang. According to §6.6.3/2 [stmt.return],
A return statement with a braced-init-list initializes the object or reference to be returned from the function by copy-list-initialization (8.5.4) from the specified initializer list.
So, there is an invisible `=` sign there and you can't get around it.
Problem
In the following code I am not allowed to declare an explicit ctor because the compiler says I am using it in a copy-initializing context (clang 3.3 and gcc 4.8). I try to prove the compilers wrong by making the ctor non explicit and then declaring the copy constructors as deleted. Are the compilers wrong or is there any other explanation? ``` #include <iostream> template <typename T> struct xyz { constexpr xyz (xyz const &) = delete; constexpr xyz (xyz &&) = delete; xyz & operator = (xyz const &) = delete; xyz & operator = (xyz &&) = delete; T i; /*explicit*/ constexpr xyz (T i): i(i) { } }; template <typename T> xyz<T> make_xyz (T && i) { return {std::forward<T>(i)}; } int main () { //auto && x = make_xyz(7); auto && x (make_xyz(7)); // compiler sees copy-initialization here too std::cout << x.i << std::endl; } ``` Update An unrealistic but much simpler version ``` struct xyz { constexpr xyz (xyz const &) = delete; constexpr xyz (xyz &&) = delete; xyz & operator = (xyz const &) = delete; xyz & operator = (xyz &&) = delete; int i; explicit constexpr xyz (int i): i(i) { } }; xyz make_xyz (int && i) { return {i}; } int main () { xyz && x = make_xyz(7); } ```