c++ 为什么在返回从函数的返回类型派生的类型的本地对象时,没有选择move构造函数？

(中继版本)拒绝：

#include <memory>

struct Base 
{
    Base() = default; 
    Base(Base const&) = delete;
    Base(Base&&) = default;
};

struct Derived : Base
{
    Derived() = default; 
    Derived(Derived const&) = delete;
    Derived(Derived&&) = default;
};    

auto foo()
    -> Base
{
    Derived d;    
    return d;   // ERROR HERE
}

根据[class.copy] / 32：

When the criteria for elision of a copy/move operation are met, but not for an exception-declaration, and the object to be copied is designated by an lvalue, or when the expression in a return statement is a (possibly parenthesized) id-expression that names an object with automatic storage duration declared in the body or parameter-declaration-clause of the innermost enclosing function or lambda-expression, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue

如果上面的句子意图被解析为(复制elision标准符合& lvalue)|| (id-expression指定一个自动对象),正如 this CWG defect 所示,为什么这不是最后一个条件呢？在Clang和GCC中是否有编译器错误？

另一方面,如果这个句子意图被解析为(copy elision criteria met&(lvalue || id-expression指定一个自动对象)),这不是一个非常误导性的字面意义吗？

[class.copy] / 32继续：

[…] if the type of the first parameter of the selected constructor  is not an rvalue reference to the object’s type (possibly  cv-qualified), overload resolution is performed again, considering the  object as an lvalue.

第一个重载分辨率,以d作为值,选择Base :: Base(Base&&).然而,所选构造函数的第一个参数的类型不是派生和&但是Base&&&&&&&&&&&&&&&第二个重载分辨率选择删除的拷贝构造函数.