c++ - 编译器的大括号初始化器列表与类型 `std::initializer_list` 的耦合程度如何？

Question

我可以在没有 C++ 头文件的情况下实现相同的效果吗 <initializer_list> ？

是否 class initializer_list必须住在namespace std (编译器是否需要这个)？

我很满意适用于五巨头(GCC、MSVC、Intel、Clang、Comeau)的解决方案

Answer 1

它按照标准紧密耦合。它是编译器构造的数组的包装器(尽管标准没有说明精确度——GCC 使用带指针和大小的构造函数，MSVC 使用带两点的构造函数)。是的，它必须在命名空间 std 中，并在 header 中 <initializer_list> .

您必须查看它们在您想要支持的每个编译器中使用的构造函数，并将其复制到您的版本中，但是老实说，我没有明白这一点。

The template std::initializer_list is not predefined; if the header <initializer_list> is not included prior to a use of std::initializer_list — even an implicit use in which the type is not named (7.1.6.4) — the program is ill-formed.

An object of type std::initializer_list<E> is constructed from an initializer list as if the implementation allocated an array of N elements of type E, where N is the number of elements in the initializer list. Each element of that array is copy-initialized with the corresponding element of the initializer list, and the std::initializer_list<E> object is constructed to refer to that array.