How 'undefined' a race condition can be?

问题

Let's say I define a following C++ object:

class AClass
{
public:
    AClass() : foo(0) {}
    uint32_t getFoo() { return foo; }
    void changeFoo() { foo = 5; }
private:
    uint32_t foo;
} aObject;

The object is shared by two threads, T1 and T2. T1 is constantly calling getFoo() in a loop to obtain a number (which will be always 0 if changeFoo() was not called before). At some point, T2 calls changeFoo() to change it (without any thread synchronization).

Is there any practical chance that the values ever obtained by T1 will be different than 0 or 5 with modern computer architectures and compilers? All the assembler code I investigated so far was using 32-bit memory reads and writes, which seems to save the integrity of the operation.

What about other primitive types?

Practical means that you can give an example of an existing architecture or a standard-compliant compiler where this (or a similar situation with a different code) is theoretically possible. I leave the word modern a bit subjective.

---

Edit: I can see many people noticing that I should not expect 5 to be read ever. That is perfectly fine to me and I did not say I do (though thanks for pointing this aspect out). My question was more about what kind of data integrity violation can happen with the above code.

回答1:

In practice, all mainstream 32-bit architectures perform 32-bit reads and writes atomically. You'll never see anything other than 0 or 5.

回答2:

In practice, you will not see anything else than 0 or 5 as far as I know (maybe some weird 16 bits architecture with 32 bits int where this is not the case).

However whether you actually see 5 at all is not guaranteed.

Suppose I am the compiler.

I see:

while (aObject.getFoo() == 0) {
    printf("Sleeping");
    sleep(1);
}

I know that:

• printf cannot change aObject
• sleep cannot change aObject
• getFoo does not change aObject (thanks inline definition)

And therefore I can safely transform the code:

while (true) {
    printf("Sleeping");
    sleep(1);
}

Because there is no-one else accessing aObject during this loop, according to the C++ Standard.

That is what undefined behavior means: blown up expectations.

回答3:

Not sure what you're looking for. On most modern architectures, there is a very distinct possibility that getFoo() always returns 0, even after changeFoo has been called. With just about any decent compiler, it's almost guaranteed that getFoo(), will always return the same value, regardless of any calls to changeFoo, if it is called in a tight loop.

Of course, in any real program, there will be other reads and writes, which will be totally unsynchronized with regards to the changes in foo.

And finally, there are 16 bit processors, and there may also be a possibility with some compilers that the uint32_t isn't aligned, so that the accesses won't be atomic. (Of course, you're only changing bits in one of the bytes, so this might not be an issue.)

回答4:

In practice (for those who did not read the question), any potential problem boils down to whether or not a store operation for an unsigned int is an atomic operation which, on most (if not all) machines you will likely write code for, it will be.

Note that this is not stated by the standard; it is specific to the architecture you are targeting. I cannot envision a scenario in which a calling thread will red anything other than 0 or 5.

As to the title... I am unaware of varying degrees of "undefined behavior". UB is UB, it is a binary state.

回答5:

Is there any practical chance that the values ever obtained by T1 will be different than 0 or 5 with modern computer architectures and compilers? What about other primitive types?

Sure - there is no guarantee that the entire data will be written and read in an atomic manner. In practice, you may end up with a read which occurred during a partial write. What may be interrupted, and when that happens depends on several variables. So in practice, the results could easily vary as size and alignment of types vary. Naturally, that variance may also be introduced as your program moves from platform to platform and as ABIs change. Furthermore, observable results may vary as optimizations are added and other types/abstractions are introduced. A compiler is free to optimize away much of your program; perhaps completely, depending of the scope of the instance (yet another variable which is not considered in the OP).

Beyond optimizers, compilers, and hardware specific pipelines: The kernel can even affect the manner in which this memory region is handled. Does your program Guarantee where the memory of each object resides? Probably not. Your object's memory may exist on separate virtual memory pages -- what steps does your program take to ensure the memory is read and written in a consistent manner for all platforms/kernels? (none, apparently)

In short: If you cannot play by the rules defined by the abstract machine, you should not use the interface of said abstract machine (e.g. you should just understand and use assembly if the specification of C++'s abstract machine is truly inadequate for your needs -- highly improbable).

All the assembler code I investigated so far was using 32-bit memory reads and writes, which seems to save the integrity of the operation.

That's a very shallow definition of "integrity". All you have is (pseudo-)sequential consistency. As well, the compiler needs only to behave as if in such a scenario -- which is far from strict consistency. The shallow expectation means that even if the compiler actually made no breaking optimization and performed reads and writes in accordance with some ideal or intention, that the result would be practically useless -- your program would observe changes typically 'long' after its occurrence.

The subject remains irrelevant, given what specifically you can Guarantee.

回答6:

Undefined behavior means that the compiler can do what ever he wants. He could basically change your program to do what ever he likes, e.g. order a pizza.

See, @Matthieu M. answer for a less sarcastic version than this one. I won't delete this as I think the comments are important for the discussion.

回答7:

Undefined behavior is guaranteed to be as undefined as the word undefined.
Technically, the observable behavior is pointless because it is simply undefined behavior, the compiler is not needed to show you any particular behavior. It may work as you think it should or not or may burn your computer, anything and everything is possible.

来源：https://stackoverflow.com/questions/15020171/how-undefined-a-race-condition-can-be