That's what I wanted to do originally - use a per-object mutext.
Unfortunately the _C_mutex member in rw::__rw_synchronized is static:
That's the wrong __rw_synchronized -- this one is for non-MT builds.
The one we get in MT builds in on line 370.
struct __rw_synchronized
{
// static so that it takes up no space
static _RWSTD_EXPORT __rw_mutex _C_mutex;
...
and __rw::rw_guard doesn't have an appropriate constructor.
Intel C++ complains about it too:
/src/steleman/programming/stdcxx-intel/stdcxx-4.2.1-thread-safe/include/loc/_numpunct.h(181):
error: no instance of constructor "__rw::__rw_guard::__rw_guard"
matches the argument list
argument types are: (const __rw::__rw_mutex)
_RWSTD_MT_GUARD (_C_mutex);
^
That's because the function is const and the guard takes a non-const
reference. This might be a defect in __rw_synchronized: the mutex
member should probably be mutable.
...
And even so, this is still not thread-safe:
Two different threads [ T1 and T2 ], seeking two different locales
[en_US.UTF-8 and ja_JP.UTF-8], enter std::numpunct<_CharT>::grouping()
at the same time - because they are running on two different cores.
Those threads won't have the same numpunct facet. A single facet
is always associated with exactly one locale and the relationship
can never change. The locale class guarantees this.
They both test for
if (!(_C_flags& _RW::__rw_gr))
and then -- assuming the expression above evaluates to true -- one of
them wins the mutex [T1], and the other one [T2] blocks on the mutex.
When T1 is done and exits the function, the grouping is set to
en_US.UTF-8 and the mutex is released. Now T2 acquires the mutex, and
proceeds to setting grouping to ja-JP.UTF-8. Woe on the caller running
from T1 who now thinks he got en_US.UTF-8, but instead he gets
ja_JP.UTF-8, which was duly set so by T2, but T1 had no clue about it
(remember, the std::string grouping _charT buffer is shared by the
caller from T1 and the caller from T2).
So at a minimum, the locking must happen before evaluating the
if (!(_C_flags& _RW::__rw_gr))
expression.
You're right that there is a race here. But the race is benign
because the T2 will assign the same value to the string as T1
did (because the grouping must be the same in the same locale).
This doesn't reallocate the string but simply overwrites each
byte with its own value. On all architectures we support this
is atomic and safe.
Martin
This still doesn't solve what ends up being returned in grouping. If
we lock at the top of the function, then, when T2 acquires the mutex,
the test expression will evaluate to false. Therefore T2 will return
whatever is in grouping right now, which happens to be en_US.UTF-8 as
set by T1, when T2 really wanted ja_JP.UTF-8.
I really think the appropriate fix here -- which would address the
performance implications -- is more complex than this. I am thinking
about creating and using a (non-publicly accessible) internal locale
cache:
typedef std::map<std::string, std::locale> locale_cache;
where all the locales are stored fully initialized, on demand. There
is only one locale instantiation and initialization overhead cost per
locale. After a locale has been instantiated and placed into the
cache, the caller of any specfic locale gets a copy from the cache,
fully instantiated and initialized. But this breaks ABI, so I'm
thinking it's for stdcxx 5.
Thoughts?
--Stefan
