[Voted into the WP at the March, 2011 meeting.]

Proposed resolution (January, 2011):

Add the following new paragraph after 9.2.4 [dcl.typedef] paragraph 4:

If a typedef specifier is used to redefine in a given scope an entity that can be referenced using an elaborated-type-specifier, the entity can continue to be referenced by an elaborated-type-specifier or as an enumeration or class name in an enumeration or class definition respectively. [Example:

  struct S;
  typedef struct S S;
  int main() {
    struct S* p; // OK
  }
  struct S {};   // OK

—end example]

Notes from the April, 2005 meeting:

The CWG agreed with John Spicer's approach, i.e., permitting a typedef-name to be used in an elaborated-type-specifier only if it is declared in the same scope as the class or enumeration it names.

Note (March, 2005):

Matt Austern: Is this legal?

    struct A { };
    typedef struct A A;
    struct A* p;

Am I right in reading the standard [to say that this is ill-formed]? On the one hand it's a nice uniform rule. On the other hand, it seems likely to confuse users. Most people are probably used to thinking that 'typedef struct A A' is a null operation, and, if this code really is illegal, it would seem to be a gratuitous C/C++ incompatibility.

Mike Miller: I think you're right. 9.2.4 [dcl.typedef] paragraph 1:

A name declared with the typedef specifier becomes a typedef-name.

9.2.4 [dcl.typedef] paragraph 2:

In a given non-class scope, a typedef specifier can be used to redefine the name of any type declared in that scope to refer to the type to which it already refers.

After the typedef declaration in the example, the name X has been “redefined” — it is no longer just a class-name, it has been “redefined” to be a typedef-name (that, by virtue of the fact that it refers to a class type, is also a class-name).

John Spicer: In C, and originally in C++, an elaborated-type-specifier did not consider typedef names, so “struct X* x” would find the class and not the typedef.

When C++ was changed to make typedefs visible to elaborated-type-specifier lookups, I believe this issue was overlooked and inadvertantly made ill-formed.

I suspect we need add text saying that if a given scope contains both a class/enum and a typedef, that an elaborated type specifier lookup finds the class/enum.

Mike Miller: I'm a little uncomfortable with this approach. The model we have for declaring a typedef in the same scope as a class/enum is redefinition, not hiding (like the “struct stat” hack). This approach seems to assume that the typedef hides the class/enum, which can then be found by an elaborated-type-specifier, just as if it were hidden by a variable, function, or enumerator.

Also, this approach reduces but doesn't eliminate the incompatibility with C. For example:

    struct S { };
    {
        typedef struct S S;
        struct S* p;        // still ill-formed
    }

My preference would be for something following the basic principle that declaring a typedef-name T in a scope where T already names the type designated by the typedef should have no effect on whether an elaborated-type-specifier in that or a nested scope is well-formed or not. Another way of saying that is that a typedef-name that designates a same-named class or enumeration in the same or a containing scope is transparent with respect to elaborated-type-specifiers.

John Spicer: This strikes me as being a rather complicated solution. When we made the change to make typedefs visible to elaborated-type-specifiers we did so knowing it would make some C cases ill-formed, so this does not bother me. We've lived with the C incompatibility for many years now, so I don't personally feel a need to undo it. I also don't like the fact that you have to essentially do the old-style elaborated-type-specifier lookup to check the result of the lookup that found the typedef.

I continue to prefer the direction I described earlier where if a given scope contains both a class/enum and a typedef, that an elaborated-type-specifier lookup finds the class/enum.

Here's an example:

  typedef struct S { ... } S;
  void fs(S *x) { ... }

The big question is, to what declaration does the reference to identifier S actually refer? Is it the S that's declared as a typedef name, or the S that's declared as a class name (or in C terms, as a struct tag)? (In either case, there's clearly only one type to which it could refer, since a typedef declaration does not introduce a new type. But the debugger apparently cares about more than just the identity of the type.)

Here's a classical, closely related example:

  struct stat { ... };
  int stat();
  ... stat( ... ) ...

Does the identifier stat refer to the class or the function? Obviously, in C, you can't refer to the struct tag without using the struct keyword, because it is in a different name space, so the reference must be to the function. In C++, the reference is also to the function, but for a completely different reason.

Now in C, typedef names and function names are in the same name space, so the natural extrapolation would be that, in the first example, S refers to the typedef declaration, as it would in C. But C++ is not C. For the purposes of this discussion, there are two important differences between C and C++

The first difference is that, in C++, typedef names and class names are not in separate name spaces. On the other hand, according to section _N4868_.6.4.10 [basic.scope.hiding] (Name hiding), paragraph 2:

A class name (9.1) or enumeration name (7.2) can be hidden by the name of an object, function, or enumerator declared in the same scope. If a class or enumeration name and an object, function, or enumerator are declared in the same scope (in any order) with the same name, the class or enumeration name is hidden wherever the object, function, or enumerator name is visible.

Please consider carefully the phrase I have highlighted, and the fact that a typedef name is not the name of an object, function or enumerator. As a result, this example:

  struct stat { ... };
  typedef int stat;

Which would be perfectly legal in C, is disallowed in C++, both implicitly (see the above quote) and explicitly (see section 9.2.4 [dcl.typedef] (The typedef specifier), paragraph 3):

In a given scope, a typedef specifier shall not be used to redefine the name of any type declared in that scope to refer to a different type. Similarly, in a given scope, a class or enumeration shall not be declared with the same name as a typedef-name that is declared in that scope and refers to a type other than the class or enumeration itself.

From which we can conclude that in C++ typedef names do not hide class names declared in the same scope. If they did, the above example would be legal.

The second difference is that, in C++, a typedef name that refers to a class is a class-name; see 9.2.4 [dcl.typedef] paragraph 4:

A typedef-name that names a class is a class-name(9.1). If a typedef-name is used following the class-key in an elaborated-type-specifier (7.1.5.3) or in the class-head of a class declaration (9), or is used as the identifier in the declarator for a constructor or destructor declaration (12.1, 12.4), the program is ill-formed.

This implies, for instance, that a typedef-name referring to a class can be used in a nested-name-specifier (i.e. before :: in a qualified name) or following ~ to refer to a destructor. Note that using a typedef-name as a class-name in an elaborated-type-specifier is not allowed. For example:

  struct X { };
  typedef struct X X2;
  X x; // legal
  X2 x2; // legal
  struct X sx; // legal
  struct X2 sx2; // illegal

The final relevant piece of the standard is 9.2.4 [dcl.typedef] paragraph 2:

In a given scope, a typedef specifier can be used to redefine the name of any type declared in that scope to refer to the type to which it already refers.

This of course is what allows the original example, to which let us now return:

  typedef struct S { ... } S;
  void fs(S *x) { ... }

The question, again is, to which declaration of S does the reference actually refer? In C, it would clearly be to the second, since the first would be accessible only by using the struct keyword. In C++, if typedef names hid class names declared in the same scope, the answer would be the same. But we've already seen that typedef names do not hide class names declared in the same scope.

So to which declaration does the reference to S refer? The answer is that it doesn't matter. The second declaration of S, which appears to be a declaration of a typedef name, is actually a declaration of a class name (9.2.4 [dcl.typedef] paragraph 4), and as such is simply a redeclaration. Consider the following example:

  typedef int I, I;
  extern int x, x;
  void f(), f();

To which declaration would a reference to I, x or f refer? It doesn't matter, because the second declaration of each is really just a redeclaration of the thing declared in the first declaration. So to save time, effort and complexity, the second declaration of each doesn't add any entry to the compiler's symbol table.