T v {}初始化 [英] T v{} initialization

问题描述

我正在阅读C ++ 11标准，但无法确定

I'm reading the C++11 standard, but can't figure out whether

T x{};

是值初始化或默认初始化（自动存储）。
表示：

is value-initialized or default initialized (automatic storage). It does say pretty clearly that:

10初始值为空的对象括号，即（），应进行值初始化。

10 An object whose initializer is an empty set of parentheses, i.e., (), shall be value-initialized.

而

11如果没有为对象指定初始化器，则对象将被默认初始化;

11 If no initializer is specified for an object, the object is default-initialized;

但我能找到 T x {}; 是：

初始化发生在形式T x（a）;
T x {a};
以及新表达式（5.3.4），static_cast表达式（5.2.9），功能符号类型转换（5.2.3）和基本和成员初始化器（12.6.2）被称为直接初始化。

The initialization that occurs in the forms T x(a); T x{a}; as well as in new expressions (5.3.4), static_cast expressions (5.2.9), functional notation type conversions (5.2.3), and base and member initializers (12.6.2) is called direct-initialization.

和

（非圆括号）支撑初始化列表，对象或引用是列表初始化的（8.5.4）。

If the initializer is a (non-parenthesized) braced-init-list, the object or reference is list-initialized (8.5.4).

新的潜水到阅读标准的水平。

I'm new to diving into the level of reading the standards. Can someone point me in the right direction?

推荐答案

这是您的报价确实涵盖的：

This is indeed covered by your quote:

如果初始化程序是（非括号） braced-init-list ，则对象或引用将被列表初始化（8.5.4）

If the initializer is a (non-parenthesized) braced-init-list, the object or reference is list-initialized (8.5.4).

跳到8.5.4列表初始化。这里我改写/省略了一些与 T x {} 的情况无关的点：

Skipping down to 8.5.4 List-initialization. Here I have paraphrased/omitted some points that don't pertain to the case of T x{}:

类型T的对象或引用的列表初始化定义如下：

List-initialization of an object or reference of type T is defined as follows:

• 如果T是聚集


• 否则，如果初始化器列表没有元素，T是带有默认构造函数的类类型，则对象将被初始化为
li>

• 否则，如果 T 是 std :: initializer_list< E> [...]


• 否则，[如果列表不为空并且它与构造函数匹配]


• 否则，[如果列表有单元素]


• 否则，[if T 是引用类型]


• 如果初始化器列表没有元素，则该对象是值初始化的。


• 否则，程序是错误的。

• If T is an aggregate, aggregate initialization is performed (8.5.1).
• Otherwise, if the initializer list has no elements and T is a class type with a default constructor, the object is value-initialized
• Otherwise, if T is a specialization of std::initializer_list<E> [...]
• Otherwise, [if the list is not empty and it matches a constructor]
• Otherwise, [if the list has a single element]
• Otherwise, [if T is a reference type]
• Otherwise, if the initializer list has no elements, the object is value-initialized.
• Otherwise, the program is ill-formed.

第一点，聚集初始化也在C ++ 03中;在这种情况下 T x {}; 与 T x = {}; 相同。

The first point, aggregate initialization was in C++03 as well; in that case T x{}; is the same as T x = {};.

对于第二点T是具有默认构造函数的类类型，它是值初始化，意味着调用默认构造函数。

For the second point "T is a class type with a default constructor", it is value-initialized which means calling the default constructor.

如果 T 是原始类型，则第二个到最后一个点适用，并且它是 >。

If T is a primitive type then the second-to-last point applies and it is value-initialized again.

返回到集合初始化情况，在8.5.1 / 7中有：

Going back to the aggregate initialization case, in 8.5.1/7 there is:

如果列表中的初始化子句数少于聚合中的成员数，则未明确初始化的每个成员都应从它的大括号或初始值初始化或者，如果没有括号或等号初始化器，则从空初始化器列表（8.5.4）。

If there are fewer initializer-clauses in the list than there are members in the aggregate, then each member not explicitly initialized shall be initialized from its brace-or-equal-initializer or, if there is no brace-or-equal-initializer, from an empty initializer list (8.5.4).

括号或初始化器是指类定义中内联的初始化器。如果不存在，那么它被初始化，如同成员已经用 {} 初始化了（所以，这个逻辑被递归地应用于每个聚合成员）。

The brace-or-equal-initializer refers to an initializer provided inline in the class definition. If that isn't present then it is initialized as if the member had been initialized with {} (so, this logic is recursively applied for each aggregate member).

例如，

struct T
{
     int a;
};

然后 T x {}; a 被初始化为 int a {}; ，这是值初始化，因为 int 是原始类型。

then T x {}; leads to a being initialized as if it were int a{}; , which is value-initialization since int is a primitive type.

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