螺纹安全和静态功能

Question

Let's suppose I have a :

class base
{
    base(){f(this);};

    static void f(base * b) {(b->d)++;};

    int d;
};

Now if on 2 separate threads I create an object of type base, would method f be considered thread safe? I am asking this question because usually from what I know is that for a method to be thread safe it should not use static members nor global variables. But as you can see from the above example I decided not to make variable d static, instead I call it through the running instance of base.

Also, I think I don't need to protect this line : (b->d)++; with a mutex since each thread will have a separate instance of base and of variable d.

Am I correct in my analysis? is there anything I should be careful about?

Answer 1

Yes, your constructor is thread safe, because it accesses only instance variables (specifically, d ). It does exhibit undefined behavior, because it reads from uninitialized d to perform the increment, but that has nothing to do with thread safety.

Here is how you can fix undefined behavior:

base(): d(0) {f(this);};

Now that d is initialized in the initializer list, your program behaves in a predictable way.