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不选择 move 构造函数？

Question

以下代码均被 Clang 拒绝和 GCC (主干版本):

#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
}

导致以下错误:

prog.cc: In function 'Base foo()': prog.cc:21:12: error: use of deleted function 'Base::Base(const Base&)'
     return d;
            ^

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

如果上面的句子被解析为 (复制省略条件满足 && lvalue) || (id-expression 指定一个自动对象)，如 this CWG defect似乎表明，为什么最后一个条件在这里不适用？ Clang 和 GCC 中是否存在编译器错误？

另一方面，如果句子被解析为(复制省略条件满足&&(左值|| id-expression指定一个自动对象))，这不是很误导性的措辞值得 DR？

Answer 1

[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.

第一个重载决议，将 d 视为右值，选择 Base::Base(Base&&)。然而，所选构造函数的第一个参数的类型不是 Derived&& 而是 Base&&，因此该重载解析的结果被丢弃，您再次执行重载解析，将d 作为左值。第二个重载决议选择已删除的复制构造函数。