[Bug c++/12581] [3.4 Regression] gcc rejects typeof use for the return type

["gabor dot greif at lucent dot com" <gcc-bugzilla at gcc dot gnu dot org>]

------- Additional Comments From gabor dot greif at lucent dot com  2003-11-12 14:03 -------
(In reply to comment #6)
> ...
> 
> | I ask myself why a typeof construct cannot be an unqualified-id?
> 
> Meaning?
> 
> -- Gaby
> 

Hmmm, I assume you do not understand what I am meaning. (Your question is a bit
overloaded :-)

Here is what I think. Let's modify my the simplified example:



struct bar
{
    void setfoobar(int);
    int getfoobar(void) const;
    static int (bar::* const getter1)(void) const;
};

template <typename T>
T* JustAPointer(void);

typedef __typeof__(JustAPointer<bar>()->getfoobar()) GETTER_RESULT;

GETTER_RESULT (bar::* const bar::getter1)(void)
const = &bar::getfoobar;
--------------------------------------------------------------------

I strongly assume this is legal _and_ compiles.
GETTER_RESULT is a typedef-name, right?
The new C++ parser only accepts an unqualified-id in that
position, so typedef-name implies unqualified-id.

Now, by statement of Mark (comment #3), a typeof construct can be used
everywhere a typedef-name can occur. So typeof-construct implies typedef-name
(basically), and by transitiveness: typeof-construct implies unqualified-id.

Maybe I am a bit dense today, but I still cannot see what is illegal in the
example in comment #1.

-- 


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=12581