Thread Safety and static functions

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?

you are correct, as long as the base pointers are thread specific you are good. — perreal
– perreal, Commented Dec 28, 2012 at 5:22
Are you able to compile your program? It should throw up a compilation error — stamhaney
– stamhaney, Commented Dec 28, 2012 at 5:22
@stamhaney, no I haven't I am just trying to work out the theory :) please point out the problem though — Kam
– Kam, Commented Dec 28, 2012 at 5:23
Doesn't HAVE to be under public access (singletons are generally made with private/protected constructors), but for most use cases, sure. I think it's clear this is just an example, though, and the brain's compiler is much more relaxed than the CPU's. — Soup in Boots
– Soup in Boots, Commented Dec 28, 2012 at 5:55

Sergey Kalinichenko · Accepted Answer · 2012-12-28 05:22:48Z

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.

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