Can "typedef typename" be replaced by "using" keyword in C++ 11

Question

I know that the "using" keyword could be used as template alias and type alias, but I didn't see anyone has mentioned that "typedef typename" could be replaced by "using". So could it?

Answer 1

A declaration of the following form

typedef typename something<T>::type alias;

can be replaced by

using alias = typename something<T>::type;

The typename is still necessary, but it does look neater, and having that = in the middle of the line is aesthetically pleasing since we are defining an alias. The main selling point of using is that it can be templated.

template <typename T>
using alias = typename something<T>::type;

You can't do that with typedef alone. The C++98 equivalent would be this slightly monstrous syntax.

template <typename T>
struct Alias {
  typedef typename something<T>::type type;
};

// Then refer to it using typename Alias<T>::type.
// Compare to the C++11 alias<T>, which is much simpler.

The other major selling point of using over typedef is that it looks much cleaner when defining function aliases. Compare

// C++98 syntax
typedef int(*alias_name)(int, int);
// C++11 syntax
using alias_name = int(*)(int, int);

Answer 2

using can replace type declaration in case if you use it as alias of another type or if you can declare this type. Syntax is:

using identifier attr(optional) = type-id ; (1)

template < template-parameter-list > using identifier attr(optional) = type-id ; (2)

type-id - abstract declarator or any other valid type-id (which may introduce a new type, as noted in type-id). The type-id cannot directly or indirectly refer to identifier.

So, this cannot be replaced by single using, you need two:

typedef struct MyS 
{
   MyS *p;
} MyStruct, *PMyStruct;