c++ - C++1z 中的§12.3.2 [class.conv.fct]/1 相对于 C++14 发生了重大变化。是否有意义？

Question

C++14 中的§12.3.2 [class.conv.fct]/1:

A member function of a class X having no parameters with a name of the form

    conversion-function-id:
       operator conversion-type-id
    conversion-type-id:
       type-specifier-seq conversion-declarator_opt
    conversion-declarator:
       ptr-operator conversion-declarator_opt

specifies a conversion from X to the type specified by the conversion-type-id. Such functions are called conversion functions. No return type can be specified. If a conversion function is a member function, the type of the conversion function (8.3.5) is “function taking no parameter returning conversion-type-id”. ...

C++1z 中的§12.3.2 [class.conv.fct]/1:(相关更改在下面突出显示)

A member function of a class X having no parameters with a name of the form

    conversion-function-id:
       operator conversion-type-id
    conversion-type-id:
       type-specifier-seq conversion-declarator_opt
    conversion-declarator:
       ptr-operator conversion-declarator_opt

specifies a conversion from X to the type specified by the conversion-type-id. Such functions are called conversion functions. A decl-specifier in the decl-specifier-seq of a conversion function (if any) shall be neither a defining-type-specifier nor static. Type of the conversion function (8.3.5) is “function taking no parameter returning conversion-type-id”. ...

我想不出没有 decl-specifier-seq 或带有不包含 的 decl-specifier-seq 的转换函数>定义类型说明符。据我所知，这意味着 conversion-function-id，其 conversion-type-id 没有类型!

Answer 1

conversion-type-id 不包含decl-specifier-seq；它包含一个类型说明符序列。

那句话是关于转换函数声明/定义中的这个decl-specifier-seq:

struct C {
    constexpr inline explicit operator int() const { return 0; }
//  ^^^^^^^^^^^^^^^^^^^^^^^^^
    operator float() const;
// ^ no decl-specifier-seq here
};

由于显而易见的原因，它不能包含定义类型说明符。

更改的要点在于，当返回类型由 conversion-type-id 确定时，说“无法指定返回类型”会产生误导。