c++ std:: types instead of corresponding c types

Question

For instance, consider the <cstdint> header. C++ standard says:

The header defines all types and macros the same as the C standard library header <stdint.h>.

So is there a need to use say std::int8_t instead of the short C form int8_t somewhere?

Answer 1

So is there a need to use say std::int8_t instead of the short C form int8_t somewhere?

Yes. Per [headers]/4 it is unspecified if the C names are defined in the global namespace or not. Because of this you need std::int8_t to use int8_t. It doesn't guarantee that std::int8_t exists, just that if it does exist, using std::int8_t is guaranteed to work.

You can just add using std::int8_t; so you don't have to type std::int8_t all over the place.

---

The relevant part of [headers]/4 is

In the C++ standard library, however, the declarations (except for names which are defined as macros in C) are within namespace scope of the namespace std

^{emphasis mine}

This means we know they are declared in std::. The part

It is unspecified whether these names (including any overloads added in [language.support] through [thread] and [depr]) are first declared within the global namespace scope and are then injected into namespace std by explicit using-declarations ([namespace.udecl]).

then allows the names to be in the global namespace, but does required it. This lets implementations do things like

<cstdint>:

namespace std
{
    #include <stdint.h>
}

or

<cstdint>:

#include <stdint.h>
namespace std
{
    using ::int8_t;
    using ::int16_t;
    //...
}

Bot accomplish the same thing (putting the names in std::) and are legal implementations but only the second one puts the names in the global scope.