Re: Avoiding the anonymous variable declaration RAII error...

Greg Herlihy <>
Fri, 14 Mar 2008 12:00:55 CST
On Mar 13, 10:47 pm, "Eric J. Holtman" <> wrote:

Greg Herlihy <> wrote

Actually there is a very simple technique to prevent an unnamed Locker
variable - just replace "Locker" in the variable's declaration with an
equivalent typedef:

     typedef Locker StLocker;

     process_threaded_data ()
          StLocker (m_mutex); // Error
          StLocker lk (m_mutex); // OK
          // ...

At least on MS Visual Studio 2005, that does not
produce an error.

So, would you expect the Visual Studio 2005 C++ compiler to compile
this program successfully:

     struct Mutex

     struct Locker
         Locker(const Mutex&) {}

     typedef Locker StLocker;

     Mutex m_mutex;

     int main()
         StLocker (m_mutex); // ?? OK or Error?

In fact no version (6, 7, or 8) of the Visual Studio C++ compiler, nor
the gcc, Comeau, or EDG C++ compilers will successfully compile the
above program. And in each case, the C++ compiler reports the same

     Error: 'Locker' : no appropriate default constructor available

According to this error message, a C++ compiler interprets this

    StLocker (m_mutex)

as the equivalent to this declaration:

    Locker m_mutex;

In both cases, the program is declaring (and default-constructing) a
Locker object named "m_mutex" - and in neither case is the program
initializing an unnamed Locker object with a Mutex variable named
"m_mutex". So the difference between using a typedef name and the
equivalent class name in this context - has a dramatic effect on the
declaration's meaning.

So, unless Locker's Mutex initializer is in fact optional, the fact
that your program allows Locker class to be default-constructed - is a
serious flaw in Locker's implementation; especially since a typedef
name can - as shown above - easily conceal the fact that a particular
Locker object is being default-constructed.

I wouldn't expect it to: a typedef doesn't introduce
a new type, it's just a synonym.

Your premise does not support your conclusion. The fact that a typedef
name does not name a type in its own right (like a class name does) -
is exactly the reason why a typedef name does not behave exactly like
a class name in every context.


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