什么是std :: initializer_list的底层结构？ [英] What is the underlying structure of std::initializer_list?

问题描述

第一部分：

std :: initializer_list 的C ++ 11，所以我想知道如何在标准库中实现。从我在这里中读取的内容，编译器创建了 T 并且给出 initializer_list< T> 。





的指针。 code> initializer_list 将创建一个引用相同数据的新对象：为什么会这样？我会猜到它：

• 复制新 initializer_list


• 将资料的拥有权移至新 initializer_list

第二部分：

std :: vector 构造函数：

  vector（initializer_list< value_type> il，
 const allocator_type& allocator_type（））; 
 





（6）初始化列表构造函数 / p>

使用 il 中的每个元素的副本构建一个容器。

< blockquote>

我不习惯move语义，但是不能将 il 的数据移动到 vector ？我不知道 std :: vector 的深层实现，但是IIRC它使用了普通的数组。

解决方案

std :: initializer_list ？

的基础结构是什么？ / blockquote>

最有可能的，只是一对指针，或指针和大小。 C ++ 11标准的第18.9 / 2段甚至在（非规范）注释中提到这一点：




code> initializer_list< E> 提供对 const E 类型的对象数组的访问。 [注意：
一对指针或一个指针加一个长度将是 initializer_list 的明显表示。
initializer_list 用于实现8.5.4中规定的初始化列表。复制初始化程序列表执行
不复制基础元素。 -end note]

此外：

我不习惯move语义，但是不能将 il 的数据移动到向量？

否，您不能从 initializer_list 的元素移动，因为 initializer_list 应该是不可变的（参见上面引用段落的第一句）。这也是为什么只有 const - 限定成员函数可以访问这些元素的原因。

First part :

std::initializer_list is a really helpful feature of C++11, so I wondered how it is implemented in the standard library. From what I read here, the compiler creates an array of type T and gives the pointer to the initializer_list<T>.

It also states that copying an initializer_list will create a new object referencing the same data : why is it so ? I would have guessed that it either :

• copies the data for the new initializer_list
• moves ownership of the data to the new initializer_list

Second part :

From just one of many online references for the std::vector constructors:

vector (initializer_list<value_type> il,
    const allocator_type& alloc = allocator_type());

(6) initializer list constructor

Constructs a container with a copy of each of the elements in il, in the same order.

I am not comfortable with move semantics yet, but couldn't the data of il be moved to the vector ? I am not aware of the deep implementation of std::vector but IIRC it uses plain-old arrays.

解决方案

What is the underlying structure of std::initializer_list?

Most likely, just a pair of pointers, or a pointer and a size. Paragraph 18.9/2 of the C++11 Standard even mentions this in a (non-normative) note:

An object of type initializer_list<E> provides access to an array of objects of type const E. [ Note: A pair of pointers or a pointer plus a length would be obvious representations for initializer_list. initializer_list is used to implement initializer lists as specified in 8.5.4. Copying an initializer list does not copy the underlying elements. —end note ]

Moreover:

I am not comfortable with move semantics yet, but couldn't the data of il be moved to the vector?

No, you can't move from the elements of an initializer_list, since elements of an initializer_list are supposed to be immutable (see the first sentence of the paragraph quoted above). That's also the reason why only const-qualified member functions give you access to the elements.

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