没有构造函数的 C++ 零初始化

Question

我不明白对于其成员具有默认值的结构的零初始化会发生什么。

如果我有这些结构：

struct A {
  int *a;
  int b;
};

struct B {
  int *a;
  int b;
  B() : b(3) {}
};

struct C {
  int *a;
  int b = 3;
};

我们可以毫无疑问地说的是：

A a; 保留所有字段未初始化
A a{}; 是 {nullptr, 0}
B b;和B b{};都是{garbage, 3}（调用了构造函数）

现在不清楚当我执行以下操作时会发生什么，这里是使用 gcc 的结果：

C c; // {garbage, 3}
C c{}; // {nullptr, 3}

问题是：C c{}; 是否保证 C::a 被初始化为 nullptr，换句话说，是否有像这样的默认成员在 C 中，如果我像 C c{};?

这样显式构造对象，仍然会零初始化其他成员

因为如果我有一个构造函数与 C 做同样的事情（比如 B），那么其他成员是 而不是 零初始化，但为什么呢？ B 和 C 有什么区别？

Answer 1

从 C++14 开始，C 是一个聚合（类似于 A），C c{} 语法执行聚合初始化。这部分包括：

[dcl.init.aggr]/8 If there are fewer initializer-clauses in the list than there are elements in a non-union aggregate, then each element not explicitly initialized is initialized as follows:
(8.1) — If the element has a default member initializer (12.2), the element is initialized from that initializer.
(8.2) — Otherwise, if the element is not a reference, the element is copy-initialized from an empty initializer list (11.6.4).
(8.3) — Otherwise, the program is ill-formed.

所以C c{};等价于C c{{}, 3};。用空列表初始化 int* 成员会导致它被零初始化。

在 C++11 中，C 不是聚合（具有默认成员初始值设定项是不合格的），并且 C c{}; 调用隐式定义的构造函数，留下 c.a 成员未初始化。

在标准的所有版本中，由于用户定义的构造函数，B 不是聚合。 B b{}; 调用该构造函数，它显式初始化 b 成员并选择让 a 未初始化。

Answer 2

Aggregate initialization - cppreference.com

If the number of initializer clauses is less than the number of members or initializer list is completely empty, the remaining members are value-initialized. If a member of a reference type is one of these remaining members, the program is ill-formed.

(until C++11)

If the number of initializer clauses is less than the number of members and bases (since C++17) or initializer list is completely empty, the remaining members and bases (since C++17) are initialized by their default member initializers, if provided in the class definition, and otherwise (since C++14) copy-initialized from empty lists, in accordance with the usual list-initialization rules (which performs value-initialization for non-class types and non-aggregate classes with default constructors, and aggregate initialization for aggregates). If a member of a reference type is one of these remaining members, the program is ill-formed.

(since C++11)

所以A a{};所有成员都默认初始化

没有构造函数的 C++ 零初始化

C++ zero initialization without constructor

c++

constructor

struct

initialization