C++中原子变量的线程安全初始化

Question

Consider the following C++11 code where class B is instantiated and used by multiple threads. Because B modifies a shared vector, I have to lock access to it in the ctor and member function foo of B . To initialize the member variable id I use a counter that is an atomic variable because I access it from multiple threads.

struct A {
  A(size_t id, std::string const& sig) : id{id}, signature{sig} {}
private:
  size_t id;
  std::string signature;
};
namespace N {
  std::atomic<size_t> counter{0};
  typedef std::vector<A> As;
  std::vector<As> sharedResource;
  std::mutex barrier;

  struct B {
    B() : id(++counter) {
      std::lock_guard<std::mutex> lock(barrier);
      sharedResource.push_back(As{});
      sharedResource[id].push_back(A("B()", id));
    }
    void foo() {
      std::lock_guard<std::mutex> lock(barrier);
      sharedResource[id].push_back(A("foo()", id));
    }
  private:
    const size_t id;
  };
}

Unfortunately, this code contains a race condition and does not work like this (sometimes the ctor and foo() do not use the same id). If I move the initialization of id to the ctor body which is locked by a mutex, it works:

struct B {
  B() {
    std::lock_guard<std::mutex> lock(barrier);
    id = ++counter; // counter does not have to be an atomic variable and id cannot be const anymore
    sharedResource.push_back(As{});
    sharedResource[id].push_back(A("B()", id));
  }
};

Can you please help me understanding why the latter example works (is it because it does not use the same mutex?)? Is there a safe way to initialize id in the initializer list of B without locking it in the body of the ctor? My requirements are that id must be const and that the initialization of id takes place in the initializer list.

Answer 1

First, there's still a fundamental logic problem in the posted code. You use ++ counter as id . Consider the very first creation of B , in a single thread. B will have id == 1 ; after the push_back of sharedResource , you will have sharedResource.size() == 1 , and the only legal index for accessing it will be 0 .

In addition, there's a clear race condition in the code. Even if you correct the above problem (initializing id with counter ++ ), suppose that both counter and sharedResource.size() are currently 0 ; you've just initialized. Thread one enters the constructor of B , increments counter , so:

counter == 1
sharedResource.size() == 0

It is then interrupted by thread 2 (before it acquires the mutex), which also increments counter (to 2), and uses its previous value (1) as id . After the push_back in thread 2, however, we have only sharedResource.size() == 1 , and the only legal index is 0.

In practice, I would avoid two separate variables ( counter and sharedResource.size() ) which should have the same value. From experience: two things that should be the same won't be—the only time redundant information should be used is when it is used for control; ie at some point, you have an assert( id == sharedResource.size() ) , or something similar. I'd use something like:

B::B()
{
    std::lock_guard<std::mutex> lock( barrier );
    id = sharedResource.size();
    sharedResource.push_back( As() );
    //  ...
}

Or if you want to make id const:

struct B
{
    static int getNewId()
    {
        std::lock_guard<std::mutex> lock( barrier );
        int results = sharedResource.size();
        sharedResource.push_back( As() );
        return results;
    }

    B::B() : id( getNewId() )
    {
        std::lock_guard<std::mutex> lock( barrier );
        //  ...
    }
};

(Note that this requires acquiring the mutex twice. Alternatively, you could pass the additional information necessary to complete updating sharedResource to getNewId() , and have it do the whole job.)

Answer 2

When an object is being initialized, it should be owned by a single thread. Then when it is done being initialized, it is made shared.

If there is such a thing as thread-safe initialization, it means ensuring that an object has not become accessible to other threads before being initialized.

Of course, we can discuss thread-safe assignment of an atomic variable. Assignment is different from initialization.

Answer 3

You are in the sub-constructor list initializing the vector. This is not really an atomic operation. so in a multi-threaded system you could get hit by two threads at the same time. This is changing what id is. welcome to thread safety 101!

moving the initialization into the constructor surrounded by the lock makes it so only one thread can access and set the vector.

The other way to fix this would be to move this into a singelton pattern. But then you are paying for the lock every time you get the object.

Now you can get into things like double checked locking:)

http://en.wikipedia.org/wiki/Double-checked_locking