Where is size_t Defined?

Question

So I know any header from the C Compatability Headers:

Places in the global namespace each name that the corresponding cxxx header would have placed in the std namespace

I also know that these C headers were deprecated as of c++17, in favor of their compatibility "cxxx" counterparts.

Now, I believe that size_t is defined exclusively by the Standard Defines Header. So I presume this technically means that the definition of size_t in the global namespace has been deprecated?

I've been using it for years as just size_t and I'd just like a confirmation before I move to using std::size_t.

Answer 1

I presume this technically means that the definition of size_t in the global namespace has been deprecated?

Yes... but.

The Standard only mandates that std::size_t must be defined¹ by <cstddef>, it does not disallow an implementation to define ::size_t², but if the implementation does, the two definitions must match³.

As a conclusion, you should use std::size_t and should neither rely on ::size_t to be defined nor define it.

The following are UB:

// DON'T
using size_t = std::size_t;        // UB
using size_t = decltype(sizeof 1); // UB

---

¹⁾[cstddef.syn]

namespace std {
    using ptrdiff_­t = see below;
    using size_­t = see below;
    using max_­align_­t = see below;
    using nullptr_­t = decltype(nullptr);

[...]
The contents and meaning of the header <cstddef> are the same as the C standard library header <stddef.h>, except that it does not declare the type wchar_­t, that it also declares the type byte and its associated operations ([support.types.byteops]), and as noted in [support.types.nullptr] and [support.types.layout].

²⁾[extern.types]/1

For each type T from the C standard library (These types are [...] size_­t,[...].), the types ​::​T and std​::​T are reserved to the implementation[.]

³⁾[extern.types]/1

[...] when defined, ​::​T shall be identical to std​::​T.