Why is std::ssize() introduced in C++20?

c++ stl unsigned signed c++20
John Z. Li picture John Z. Li · May 20, 2019 · Viewed 8.1k times · Source

C++20 introduced the std::ssize() free function as below:

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

A possible implementation seems using static_cast, to convert the return value of the size() member function of class C into its signed counterpart.

Since the size() member function of C always returns non-negative values, why would anyone want to store them in signed variables? In case one really wants to, it is a matter of simple static_cast.

Why is std::ssize() introduced in C++20?

Answer

Nadav Har&#39;El picture Nadav Har'El · May 20, 2019

The rationale is described in this paper. A quote:

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.

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