c++11 - using vs. typedef - is there a subtle, lesser known difference?

Question

Background

Everybody agrees that

using <typedef-name> = <type>;

is equivalent to

typedef <type> <typedef-name>;

and that the former is to be preferred to the latter for various reasons (see Scott Meyers, Effective Modern C++ and various related questions on stackoverflow).

This is backed by [dcl.typedef]:

A typedef-name can also be introduced by an alias-declaration. The identifier following the using keyword becomes a typedef-name and the optional attribute-specifier-seq following the identifier appertains to that typedef-name. Such a typedef-name has the same semantics as if it were introduced by the typedef specifier.

However, consider a declaration such as

typedef struct {
    int val;
} A;

For this case, [dcl.typedef] specifies:

If the typedef declaration defines an unnamed class (or enum), the first typedef-name declared by the declaration to be that class type (or enum type) is used to denote the class type (or enum type) for linkage purposes only (3.5).

The referenced section 3.5 [basic.link] says

A name having namespace scope that has not been given internal linkage above has the same linkage as the enclosing namespace if it is the name of [...] an unnamed class defined in a typedef declaration in which the class has the typedef name for linkage purposes [...]

Assuming the typedef declaration above is done in the global namespace, the struct A would have external linkage, since the global namespace has external linkage.

Question

The question is now whether the same is true, if the typedef declaration is replaced by an alias declaration according to the common notion that they are equivalent:

using A = struct {
    int val;
};

In particular, does the type A declared via the alias declaration ("using") have the same linkage as the one declared via the typedef declaration?

Note that [decl.typedef] does not say that an alias declaration is a typedef declaration (it only says that both introduce a typedef-name) and that [decl.typedef] speaks only of a typedef declaration (not an alias declaration) having the property of introducing a typedef name for linkage purposes. If the alias declaration is not capable of introducing a typedef name for linkage purposes, A would just be an alias for an anonymous type and have no linkage at all.

IMO, that's at least one possible, albeit strict, interpretation of the standard. Of course, I may be overlooking something.

This raises the subsequent questions:

• If there is indeed this subtle difference, is it by intention or is it an oversight in the standard?
• What is the expected behavior of compilers/linkers?

Research

The following minimal program consisting of three files (we need at least two separate compilation units) is used to investigate the issue.

a.hpp

#ifndef A_HPP
#define A_HPP

#include <iosfwd>

#if USING_VS_TYPEDEF
using A = struct {
     int val;
};
#else
typedef struct {
     int val;
} A;
#endif

void print(std::ostream& os, A const& a);

#endif // A_HPP

a.cpp

#include "a.hpp"
#include <iostream>

void print(std::ostream& os, A const& a)
{
   os << a.val << "
";
}

main.cpp

#include "a.hpp"
#include <iostream>

int main()
{
    A a;
    a.val = 42;
    print(std::cout, a);
}

GCC

Compiling this with gcc 7.2 with the "typedef" variant compiles cleanly and provides the expected output:

> g++ -Wall -Wextra -pedantic-errors -DUSING_VS_TYPEDEF=0 a.cpp main.cpp
> ./a.out
42

The compilation with the "using" variant produces a compile error:

> g++ -Wall -Wextra -pedantic-errors -DUSING_VS_TYPEDEF=1 a.cpp main.cpp
a.cpp:4:6: warning: ‘void print(std::ostream&, const A&)’ defined but not used [-Wunused-function]
void print(std::ostream& os, A const& a)
     ^~~~~
In file included from main.cpp:1:0:
a.hpp:16:6: error: ‘void print(std::ostream&, const A&)’, declared using unnamed type, is used but never defined [-fpermissive]
void print(std::ostream& os, A const& a);
     ^~~~~
a.hpp:9:2: note: ‘using A = struct<unnamed>’ does not refer to the unqualified type, so it is not used for linkage
};
 ^
a.hpp:16:6: error: ‘void print(std::ostream&, const A&)’ used but never defined
void print(std::ostream& os, A const& a);
     ^~~~~

This looks like GCC follows the strict interpretation of the standard above and makes a difference concerning linkage between the typedef and the alias declaration.

Clang

Using clang 6, both variants compile and run cleanly without any warnings:

> clang++ -Wall -Wextra -pedantic-errors -DUSING_VS_TYPEDEF=0 a.cpp main.cpp
> ./a.out
42

> clang++ -Wall -Wextra -pedantic-errors -DUSING_VS_TYPEDEF=1 a.cpp main.cpp
> ./a.out
42

One could therefore also ask

• Which compiler is correct?

Answer 1

This looks to me like a bug in GCC.

Note that [decl.typedef] does not say that an alias declaration is a typedef declaration

You're right, [dcl.dcl]p9 gives a definition of the term typedef declaration which excludes alias-declarations. However, [dcl.typedef] does explicitly say, as you quoted in your question:

2 A typedef-name can also be introduced by an alias-declaration. The identifier following the using keyword becomes a typedef-name and the optional attribute-specifier-seq following the identifier appertains to that typedef-name. It has the same semantics as if it were introduced by the typedef specifier. [...]

"The same semantics" doesn't leave any doubt. Under GCC's interpretation, typedef and using have different semantics, therefore the only reasonable conclusion is that GCC's interpretation is wrong. Any rules applying to typedef declarations must be interpreted as applying to alias-declarations as well.