c++ - C++20 中的指定初始值设定项

Question

我有一个关于 c++20 功能之一的问题，即指定初始值设定项(有关此功能的更多信息 here )

#include <iostream>

constexpr unsigned DEFAULT_SALARY {10000};

struct Person
{
    std::string name{};
    std::string surname{};
    unsigned age{};
};

struct Employee : Person
{
    unsigned salary{DEFAULT_SALARY};
};

int main()
{
    std::cout << std::boolalpha << std::is_aggregate_v<Person> << '\n'; // true is printed
    std::cout << std::boolalpha << std::is_aggregate_v<Employee> << '\n'; // true is printed

    Person p{.name{"John"}, .surname{"Wick"}, .age{40}}; // it's ok
    Employee e1{.name{"John"}, .surname{"Wick"}, .age{40}, .salary{50000}}; // doesn't compile, WHY ?

    // For e2 compiler prints a warning "missing initializer for member 'Employee::<anonymous>' [-Wmissing-field-initializers]"
    Employee e2 {.salary{55000}}; 
}

此代码是使用 gcc 9.2.0 和 -Wall -Wextra -std=gnu++2a 编译的旗帜。

正如您在上面看到的，两个结构 Person和Employee是聚合，但初始化 Employee使用指定的初始值设定项不可能进行聚合。

有人可以解释一下为什么吗？

Answer 1

根据 C++ 20 标准(9.3.1 聚合。第 #3 页)

(3.1) — If the initializer list is a designated-initializer-list, the aggregate shall be of class type, the identifier in each designator shall name a direct non-static data member of the class, and the explicitly initialized elements of the aggregate are the elements that are, or contain, those members.

因此您不能使用指定的初始化器列表来初始化基类的数据成员。

使用通常的列表初始化，例如

Employee e1{ "John", "Wick", 40, 50000 };

或

Employee e1{ { "John", "Wick", 40 }, 50000 };

或者正如@Jarod42在评论中指出的那样，您可以编写

Employee e1{ { .name{"John"}, .surname{"Wick"}, .age{40} }, 50000 };

在这种情况下，直接基类由指定的初始化器列表初始化，而整个类 Employe 由非指定的初始化器列表初始化。