具有u8，char8_t和std :: string的C ++ 20 [英] C++20 with u8, char8_t and std::string

问题描述

C ++ 11为我们带来了UTF-8字面量的u8前缀，我认为几年前这很酷，并在代码中添加了以下内容：

C++11 brought us the u8 prefix for UTF-8 literals and I thought that was pretty cool a few years ago and peppered my code with things like this:

std::string myString = u8"●";

这一切都很好，但是在C ++ 20中却出现了问题似乎可以编译了，因为u8创建了char8_t *，而这与仅使用char的std :: string不兼容。

This is all fine and good, but the issue comes up in C++20 it doesn't seem to compile anymore because u8 creates a char8_t* and this is incompatible with std::string which just uses char.

我应该创建一个新的utf8string吗？在C ++ 20世界中，如果我们有更显式的类型与标准std :: string真正不匹配的显式类型，一致而正确的方法是什么？

Should I be creating a new utf8string? What's the consistent and correct way to do this kind of thing in a C++20 world where we have more explicit types that don't really match with the standard std::string?

推荐答案

除了@lubgr的答案，论文 char8_t向后兼容修复（P1423）讨论了几种使用<$ c制作 std :: string 的方法。 $ c> char8_t 字符数组。

In addition to @lubgr's answer, the paper char8_t backward compatibility remediation (P1423) discusses several ways how to make std::string with char8_t character arrays.

基本上，您可以将 u8 char数组转换为普通 char数组以获取与C ++ 17和之前的行为相同，您只需要更加明确即可。本文讨论了各种方法。

Basically the idea is that you can cast the u8 char array into a "normal" char array to get the same behaviour as C++17 and before, you just have to be a bit more explicit. The paper discusses various ways to do this.

适合用例的最简单的方法（但开销不为零，除非添加更多的重载）可能是最后一种方法，即引入显式转换函数：

The most simple (but not fully zero overhead, unless you add more overloads) method that fits your usecase is probably the last one, i.e. introduce explicit conversion functions:

std::string from_u8string(const std::string &s) {
  return s;
}
std::string from_u8string(std::string &&s) {
  return std::move(s);
}
#if defined(__cpp_lib_char8_t)
std::string from_u8string(const std::u8string &s) {
  return std::string(s.begin(), s.end());
}
#endif

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