Re: Elaborated type specifier as template argument

On 19 Aug., 02:01, GMan <gmanni...@gmail.com> wrote:


For the following discussion I read this as:

template <typename T> struct foo {};


It depends on the context, in which both typedef
occur. If both are in the same namespace (including
the global namespace), they should refer to the same
type. I read this from [basic.scope.pdecl]/6 of the current
draft (It should apply to C++03 as well, the corresponding
paragraph is [basic.scope.pdecl]/5):

The point of declaration of a class first declared in an elaborated-
type-
specifier is as follows:
? for a declaration of the form
  class-key attribute-specifieropt identifier ;
the identifier is declared to be a class-name in the scope that
contains
the declaration, otherwise
? for an elaborated-type-specifier of the form
  class-key identifier
if the elaborated-type-specifier is used in the decl-specifier-seq or
parameter-declaration-clause of a function defined in namespace
scope,
the identifier is declared as a class-name in the namespace that
contains the declaration; otherwise, except as a friend declaration,
the
identifier is declared in the smallest non-class, non-function-
prototype
scope that contains the declaration. [ Note: these rules also apply
within templates. ?end note ] [ Note: other forms of elaborated-type-
specifier do not declare a new name, and therefore must refer to an
existing type-name. See 3.4.4 and 7.1.6.3. ?end note ]

You example belongs to bullet 2 and falls into the "otherwise" slot.
The effects of the wording are as if the above typedefs were
rewritten as

// TU1
struct incomplete_type;
typedef foo<incomplete_type> tu1;

// TU2
struct incomplete_type;
typedef foo<incomplete_type> tu2;

HTH & Greetings from Bremen,

Daniel Kr?gler

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