#include <iostream>
struct D;
struct S{ operator D(); };

struct D{
    D(){}
    D(D&&) { std::cout << "move constructor" << '\n'; }
};

S::operator D() { std::cout << "conversion function" << '\n'; return D(); }

int main()
{
    S s;
    D d(s);
}

我基于以下理由对这个省略的正确性提出异议：@H_301_5@

>这个案例在§8.5/ 16(N3337)的第一个子项目中有所涉及,它没有提及elision.@H_301_5@

If the initialization is direct-initialization,or if it is
copy-initialization where the cv-unqualified version of the source
type is the same class as,or a derived class of,the class of the
destination,constructors are considered. The applicable constructors
are enumerated (13.3.1.3),and the best one is chosen through overload
resolution (13.3). The constructor so selected is called to initialize
the object,with the initializer expression or expression-list as its
argument(s). If no constructor applies,or the overload resolution is
ambiguous,the initialization is ill-formed.@H_301_5@

>请注意,下一个子项目明确提到了省略的可能性.
>对move构造函数的调用是显式的.怎么可以省略？@H_301_5@