Re: [c++-pthreads] Initialization of local static mutex

[Mark Mitchell <mark@xxxxxxxxxxxxxxxx>]

Jean-Marc Bourguet wrote:
> Mark Mitchell wrote:
> > Roland Schwarz wrote:
>
> > > Now that you have brought up this issue I am curious which "other
> > > issues" you are refering to.
> >
> > void f() {
> >   static int i = g();
> > }
> >
> > Here, you cannot do the initialization statically; it must be done the 
> > first time that f() is called.  So, the key question is whether the 
> > initialization is thread-safe.  Does the programmer have to change 
> > that initialization to be thread-safe, using locks in f()?  Or does 
> > the compiler take care of it?
>
> I fail to see how the programmer can make the initialization thread-safe 
> if the compiler doesn't take care of it?  Lack of memory barriers could 
> make the bool indicating that the initialization is done visible, but 
> not the written value.

The programmer can't take care of it easily.

If the compiler doesn't do it for you, have to do things like:

  static int i;
  static bool initialized;

  lock_mutex();
  if (!initialized) {
    i = g();
    initialized = true;
  }
  unlock_mutex();

i.e., you have to manage the initialization manually.  And, yes, there 
are lots of problems with that, including exceptions being thrown by 
g().  So, it's a good feature for compilers to have.  Of course, since 
it makes initialization slower (and code bigger) in the single-threaded 
case, some programmers want to be able to turn off the compiler support.

--
Mark Mitchell
CodeSourcery
mark@xxxxxxxxxxxxxxxx
(650) 331-3385 x713