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.

`Now, I believe that size_t is defined exclusively by the Standard Defines Header` is not exactly correct, since many of the C compatibility library headers besides `cstddef` define it. — eerorika, Jul 24 '18 at 14:09
@François I'm not sure this is a duplicate. OP wants to know if `::size_t` is deprecated. — YSC, Jul 24 '18 at 14:15
@YSC The duplicate does address that `::size_t` is not required to be define. I would reason that it addresses the part about deprecation because the standard can't deprecate something it doesn't define. — François Andrieux, Jul 24 '18 at 14:18

score 6 · Accepted Answer · edited Jun 20 '20 at 09:12

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.

AnT stands with Russia · Answer 2 · 2018-07-24T15:27:06.677

C-style header names like <stddef.h> are deprecated. However, C++-style headers like <cstddef> are allowed to declare their names in global namespace and then redeclare the same names in namespace std through using-declarations (http://eel.is/c++draft/organization#headers-4). This approach to declaring standard names is not deprecated. And many implementations do exactly that, for which reason it is not unusual to see name size_t accessible as a name from global namespace. But this is not guaranteed. So, in portable C++ code that includes <cstddef> you should use std::size_t and never rely on ::size_t's availability.

Maxim Egorushkin · Answer 3 · 2018-07-24T15:10:24.083

1

The standard says [expr.sizeof]:

The result of sizeof and sizeof... is a constant of type std::size_t. [ Note: std::size_t is defined in the standard header <cstddef> ([cstddef.syn], [support.types.layout]). — end note ]

edited Jul 24 '18 at 15:10

answered Jul 24 '18 at 14:12

Maxim Egorushkin

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You have a similar answer in the duplicate. – François Andrieux Jul 24 '18 at 14:15
@FrançoisAndrieux This answer is different because it does not add to `std::`. – Maxim Egorushkin Jul 24 '18 at 14:17
`Based on the above, I do not see why you cannot do: using size_t = decltype(sizeof 1);` You can't do that (in global namespace). `::size_t` is a reserved name, so defining it has undefined behaviour. – eerorika Jul 24 '18 at 14:17
@user2079303 not anymore. – YSC Jul 24 '18 at 14:18
@YSC since when? It's reserved [here](http://eel.is/c++draft/reserved.names#footnote-182). Does the draft differ from the published standard? `For each type T ... the types ::T and ... are reserved to the implementation ... These types are ... size_t ...` – eerorika Jul 24 '18 at 14:20
@user2079303 Retracted that claim, it does not seem to be too popular. – Maxim Egorushkin Jul 24 '18 at 15:10

eerorika · Answer 4 · 2018-07-24T14:42:08.190

Where is size_t Defined?

::size_t is guaranteed to be defined in <stddef.h> and few other C standard library headers that are inherited by c++. It may be defined by the implementation whether anything is included or not, since it is reserved, but relying on such non guaranteed definition would not be wise.

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

Indeed. More precisely, all standard headers that are guaranteed to define ::size_t are now deprecated in C++17. Same applies to all other non-macro C library names such as ::FILE.

Where is size_t Defined?

4 Answers4

Yes... but.

Linked