C++20 中的指定初始化器

Question

我有一个关于 c++20 特性的问题，指定初始化器（关于这个特性的更多信息 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 标志编译的。

正如您在上面看到的，Person 和 Employee 两个结构都是聚合，但是 Employee 聚合的初始化无法使用指定的初始化程序。

有人能解释一下为什么吗？

Answer 1

您可能有几个来自不同基地的同名字段，

照理说，你应该提供想要的基地的名称，但似乎没有办法做到这一点。

// Invalid too:
Employee e1{.Person.name{"John"}, .Person.surname{"Wick"}, .Person.age{40}, .salary{50000}};
Employee e2{.Person{.name{"John"}, .surname{"Wick"}, .age{40}}, .salary{50000}};

此外，C++ 指定的初始化比 C:

更受限制

Note: out-of-order designated initialization, nested designated initialization, mixing of designated initializers and regular initializers, and designated initialization of arrays are all supported in the C programming language, but are not allowed in C++.

Answer 2

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.

所以你不能使用指定的初始化列表来初始化base classes的数据成员。

改用通常的列表初始化，如

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

或

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

或如 @Jarod42 指出的那样，您可以写评论

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

在这种情况下，直接基础 class 由指定的初始化列表初始化，而 class 全体雇员由非指定的初始化列表初始化。

C++20 中的指定初始化器

Designated initializers in C++20

c++

aggregate

c++20

designated-initializer