数组类型的初始化（C++）

Question

根据https://en.cppreference.com/w/cpp/language/list_initialization，列表初始化的作用之一是：

If T is an aggregate type, aggregate initialization is performed.

由于数组是聚合类型，当我初始化数组时int array[3] = {1, 2};，

我相信发生的事情是

1. 列表初始化
2. 聚合初始化
3. 复制初始化，其中 1、2 是从初始化列表的相应子句复制初始化的
4. 剩余的值是值初始化的（零初始化）

这对我来说很有意义，因为数组的值将是 {1, 2, 0}.

然而，当我继续阅读时，我注意到列表初始化的另一个影响是：

If T is an aggregate type and the initializer list has a single element of the same or derived type (possibly cv-qualified), the object is initialized from that element (by copy-initialization for copy-list-initialization, or by direct-initialization for direct-list-initialization).

那么，在 "initializer list has a single element" 处声明一个数组 int array[3] = {1}; 与当有多个元素时是不同的过程吗？ （即 int array [3] = {1, 2};）？这对我来说没有意义，但我不确定我错过了什么。

Answer 1

不会应用该效果，如引号所述，

If T is an aggregate type and the initializer list has a single element of the same or derived type (possibly cv-qualified), the object is initialized from that element (by copy-initialization for copy-list-initialization, or by direct-initialization for direct-list-initialization).

给定 int array[3] = {1};，{1} 具有 int 类型的单个元素 1，它不是数组类型 [=14= 的相同类型或派生类型].