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

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 

namespace std {

  using ::ptrdiff_t;
  using ::size_t;

}

I think that should make it clear. As long as you include 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;

}