在Visual Studio 2012中std :: atomic_flag静态初始化线程安全吗？

Question

Visual Studio 2012 does not implement the C++11 standard for thread safe static initialization ( http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2008/n2660.htm ). I have a function local static that I need to guarantee will be initialized in a thread safe way. The following is not thread safe in Visual Studio 2012:

struct MyClass
{
    int a;
    MyClass()
    {
        std::this_thread::sleep_for(std::chrono::milliseconds(100));
        a = 5;
    }
};

void foo()
{
    static MyClass instance;
    std::cout << instance.a << '\n';
}

int main()
{
    std::thread a(foo);
    std::thread b(foo);
    a.join();
    b.join();

    system("pause");
}

The output of the above program on Visual Studio 2012 will most likely be:

0
5

I need to work around this problem and I am trying to find a way to do it with function local statics only (no globals or class level statics).

My initial thought was to use a mutex, but it suffers from the same problem of static initialization thread safety. If I have a static st::mutex inside of foo it is possible that the second thread will get a copy of the mutex while it is in an invalid state.

Another option is to add an std::atomic_flag spin-lock. The question is, is std::atomic_flag initialization thread safe in Visual Studio 2012?

void foo()
{
    // is this line thread safe?
    static std::atomic_flag lock = ATOMIC_FLAG_INIT;
    // spin lock before static construction
    while (lock.test_and_set(std::memory_order_acquire));
    // construct an instance of MyClass only once
    static MyClass instance;
    // end spin lock
    lock.clear(std::memory_order_release);
    // the following is not thread safe
    std::cout << instance.a << '\n';
}

In the above code, is it possible for both threads to get past the spin lock or is it guaranteed that only one of them will? Unfortunately I can't think of an easy way to test this since I can't put something inside the atomic_flag initializer to slow it down like I can with a class. However, I want to be sure that my program won't crash once in a blue moon because I made an invalid assumption.

Answer 1

Section 6.7.4 of C++11 states that variables with static storage duration are initialized thread-safe:

If control enters the declaration concurrently while the variable is being initialized, the concurrent execution shall wait for completion of the initialization.

But neither of VC++ 2012 or 2013 Preview implement this, so yes, you'll need some protection to make your function thread-safe.

C++11 also says this about ATOMIC_FLAG_INIT , in section 29.7.4 :

The macro ATOMIC_FLAG_INIT shall be defined in such a way that it can be used to initialize an object of type atomic_flag to the clear state. For a static-duration object, that initialization shall be static.

VC++ does happen to implement this properly. ATOMIC_FLAG_INIT is 0 in VC++, and VC++ zero-initializes all statics at application start, not in the function call. So, your use of this is safe and there will be no race to initialize lock .

Test Code:

struct nontrivial
{
    nontrivial() : x(123) {}
    int x;
};

__declspec(dllexport) int next_x()
{
    static nontrivial x;
    return ++x.x;
}

__declspec(dllexport) int next_x_ts()
{
    static std::atomic_flag flag = ATOMIC_FLAG_INIT;

    while(flag.test_and_set());
    static nontrivial x;
    flag.clear();

    return ++x.x;
}

next_x :

                mov     eax, cs:dword_1400035E4
                test    al, 1                   ; checking if x has been initialized.
                jnz     short loc_140001021     ; if it has, go down to the end.
                or      eax, 1
                mov     cs:dword_1400035E4, eax ; otherwise, set it as initialized.
                mov     eax, 7Bh                 
                inc     eax                     ; /O2 is on, how'd this inc sneak in!?
                mov     cs:dword_1400035D8, eax ; init x.x to 124 and return.
                retn
loc_140001021:
                mov     eax, cs:dword_1400035D8
                inc     eax
                mov     cs:dword_1400035D8, eax
                retn

next_x_ts :

loc_140001032:
                lock bts cs:dword_1400035D4, 0  ; flag.test_and_set().
                jb      short loc_140001032     ; spin until set.
                mov     eax, cs:dword_1400035E0
                test    al, 1                   ; checking if x has been initialized.
                jnz     short loc_14000105A     ; if it has, go down to end.
                or      eax, 1                  ; otherwise, set is as initialized.
                mov     cs:dword_1400035E8, 7Bh ; init x.x with 123.
                mov     cs:dword_1400035E0, eax

loc_14000105A:
                lock btr cs:dword_1400035D4, 0  ; flag.clear().
                mov     eax, cs:dword_1400035E8
                inc     eax
                mov     cs:dword_1400035E8, eax
                retn

You can see here that next_x is definitely not thread-safe, but next_x_ts never initializes the flag variable at cs:dword_1400035D4 -- it is zero-initialized at application start, so there are no races and next_x_ts is thread-safe.