默认函数 return 可以自动吗？

Question

考虑以下 class：

class example
{
public:
    auto & operator =(const example &) = default;
    auto & operator =(example &&) = default;
};

这些声明是否合法？

Answer 1

不是语言律师回答

根据我的经验，编译器不接受默认特殊成员函数的 auto return 类型，因此我认为它们确实是标准不允许的。

我所知道的唯一例外是 C++20 默认的三向比较运算符：

#include <compare>

struct X
{
    auto operator<=>(const X&) const = default;
};

Answer 2

在 C++20 中，是¹

Let R be the declared return type of a defaulted three-way comparison operator function, and let xi be the elements of the expanded list of subobjects for an object x of type C`.

• If R is auto, then let cvi Ri be the type of the expression xi <=> xi. The operator function is defined as deleted if that expression is not usable or if Ri is not a comparison category type ([cmp.categories.pre]) for any i. The return type is deduced as the common comparison type (see below) of R0, R1, …, Rn−1.

[class.spaceship/2]

在 C++20 之前，否

A function definition of the form: attribute-specifier-seq opt decl-specifier-seq opt declarator virt-specifier-seq opt = default ; is called an explicitly-defaulted definition. A function that is explicitly defaulted shall

• be a special member function,
• have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T”, where T is the name of the member function's class) as if it had been implicitly declared, and

[dcl.fct.def.default]（强调）

1. 但只有 <=>。默认的 == 必须 return bool，赋值与以前的标准有类似的限制。

if F is an assignment operator, and the return type of T1 differs from the return type of T2 or T1's parameter type is not a reference, the program is ill-formed;

[dcl.fct.def.default]

Answer 3

auto & operator =(const example &) = default;
auto & operator =(example &&) = default;

Are those declarations considered legal?

没有

[dcl.spec.auto] ... If the declared return type of the function contains a placeholder type, the return type of the function is deduced from non-discarded return statements, if any, in the body of the function ([stmt.if]).

默认函数定义没有主体，这与引用规则冲突。无法从中推导出 return 类型，也没有规则说明在这种情况下该类型是什么。

operator<=> 有一个例外规则，指定当 auto 被使用时 return 类型将是什么，如 所示，但 operator= 没有有这样的规则。我认为没有理由不能引入这样的规则来允许默认赋值运算符中的 auto。