c++ - 为什么在 C++20 中引入了 std::ssize()？

Question

C++20引入std::ssize() free 函数如下:

template <class C>
    constexpr auto ssize(const C& c)
        -> std::common_type_t<std::ptrdiff_t,
                              std::make_signed_t<decltype(c.size())>>;

一个可能的实现似乎是使用 static_cast，将 class C 的 size() 成员函数的返回值转换为其有符号的对应的。

既然 C 的 size() 成员函数总是返回非负值，为什么有人想将它们存储在有符号变量中？万一真的想要，这只是一个简单的static_cast。

为什么在C++20中引入std::ssize()？

Answer 1

基本原理在 this paper 中描述。 .引用:

When span was adopted into C++17, it used a signed integer both as an index and a size. Partly this was to allow for the use of "-1" as a sentinel value to indicate a type whose size was not known at compile time. But having an STL container whose size() function returned a signed value was problematic, so P1089 was introduced to "fix" the problem. It received majority support, but not the 2-to-1 margin needed for consensus.

This paper, P1227, was a proposal to add non-member std::ssize and member ssize() functions. The inclusion of these would make certain code much more straightforward and allow for the avoidance of unwanted unsigned-ness in size computations. The idea was that the resistance to P1089 would decrease if ssize() were made available for all containers, both through std::ssize() and as member functions.