Does “std::size_t” make sense in C++?

Question

In some code I\'ve inherited, I see frequent use of size_t with the std namespace qualifier. For example:

std::size_t n = sizeof(          


        

Answer 1

std::size_t n = sizeof( long );

Actually, you haven't asked what specifically seems to be a bad practice int the above. Use of size_t, qualification with std namespace,...

As the C++ Standard says (18.1), size_t is a type defined in the standard header . I'd suggest to drop any thoughts and impressions about possible inheritance from C language. C++ is a separate and different language and it's better to consider it as such. It has its own standard library and all elements of C++ Standard Library are defined within namespace std. However, it is possible to use elements of C Standard Library in C++ program.

I'd consider including as a dirty hack. The C++ Standard states that the content of headers is the same or based on corresponding headers from the C Standard Library, but in number of cases, changes have been applied. In other words, it's not a direct copy & paste of C headers into C++ headers.

size_t is not a built-in type in C++. It is a type defined to specify what kind of integral type is used as a return type of sizeof() operator, because an actual return type of sizeof() is implementation defined, so the C++ Standard unifies by defining size_t.

would the following program (with no includes) be expected to compile on all C++ compilers?

size_t foo()
{
    return sizeof( long );
}

The C++ Standard says (1.4):

The names defined in the library have namespace scope (7.3). A C ++ translation unit (2.1) obtains access to these names by including the appropriate standard library header (16.2).

The size_t is a name defined within std namespace, so every program that uses this name should include corresponding header, in this case.

Next, the 3.7.3 chapter says:

However, referring to std, std::bad_alloc, and std::size_t is ill-formed unless the name has been declared by including the appropriate header.

Given that, program using size_t but not including header is ill-formed.

Answer 2

The GNU compiler headers contain something like

typedef long int __PTRDIFF_TYPE__;
typedef unsigned long int __SIZE_TYPE__;

Then stddef.h constains something like

typedef __PTRDIFF_TYPE__ ptrdiff_t;
typedef __SIZE_TYPE__ size_t;

And finally the cstddef file contains something like

#include <stddef.h>

namespace std {

  using ::ptrdiff_t;
  using ::size_t;

}

I think that should make it clear. As long as you include <cstddef> you can use either size_t or std::size_t because size_t was typedefed outside the std namespace and was then included. Effectively you have

typedef long int ptrdiff_t;
typedef unsigned long int size_t;

namespace std {

  using ::ptrdiff_t;
  using ::size_t;

}