Is memory barrier or atomic operation required in a busy-wait loop?
This important: in C++ volatile has nothing at all to do with concurrency! The purpose of volatile is to tell the compiler that it shall not optimize accesses to the affected object. It does not tell the CPU anything, primarily because the CPU would know already whether memory would be volatile or not. The purpose of volatile is effectively to deal with memory mapped I/O.
The C++ standard is very clear in section 1.10 [intro.multithread] that unsynchronized access to an object which is modified in one thread and is accessed (modified or read) in another thread is undefined behavior. The synchronization primitives avoiding undefined behavior are library components like the atomic classes or mutexes. This clause mentions volatile only in the context of signals (i.e., as volatile sigatomic_t) and in the context of forward progress (i.e., that a thread will eventually do something which has an observable effect like accessing a volatile object or doing I/O). There is no mention of volatile in conjunction with synchronization.
Thus, unsynchronized assess to a variable shared across threads leads to undefined behavior. Whether it is declared volatile or not doesn't matter to this undefined behavior.
From the Wikipedia page on memory barriers:
... Other architectures, such as the Itanium, provide separate "acquire" and "release" memory barriers which address the visibility of read-after-write operations from the point of view of a reader (sink) or writer (source) respectively.
To me this implies that Itanium requires a suitable fence to make reads/writes visible to other processors, but this may in fact just be for purposes of ordering. The question, I think, really boils down to:
Does there exist an architecture where a processor might never update its local cache if not instructed to do so? I don't know the answer, but if you posed the question in this form then someone else might. In such an architecture your code potentially goes into an infinite loop where the read of *lock always sees the same value.
In terms of general C++ legality, the one atomic test and set in your example isn't enough, since it implements only a single fence which will allow you to see the initial state of the *lock when entering the while loop but not to see when it changes (which results in undefined behavior, since you are reading a variable that is changed in another thread without synchronisation) - so the answer to your question (1.1/3) is no.
On the other hand, in practice, the answer to (1.2/2) is yes (given GCC's volatile semantics), so long as the architecture guarantees cache coherence without explicit memory fences, which is true of x86 and probably for many architectures but I can't give a definite answer on whether it is true for all architectures that GCC supports. It is however generally unwise to knowingly rely on particular behavior of code that is technically undefined behavior according to the language spec, especially if it is possible to get the same result without doing so.
Incidentally, given that memory_order_relaxed exists, there seems little reason not to use it in this case rather than try to hand-optimise by using non-atomic reads, i.e. changing the while loop in your example to:
while (atomic_load_explicit(lock, memory_order_relaxed)) { cpu_relax(); }On x86_64 for instance the atomic load becomes a regular mov instruction and the optimised assembly output is essentially the same as it would be for your original example.
- Is volatile enough here or are there any architectures or compilers where memory or compiler barrier or atomic operation is required in the while loop?
will the volatile code see the change. Yes, but not necessarily as quickly as if there was a memory barrier. At some point, some form of synchronization will occur, and the new state will be read from the variable, but there are no guarantees on how much has happened elsewhere in the code.
1.1 According to C++ standards?
From cppreference : memory_order
It is the memory model and memory order which defines the generalized hardware that the code needs to work on. For a message to pass between threads of execution, an inter-thread-happens-before relationship needs to occur. This requires either...
- A synchronizes-with B
- A has a std::atomic operation before B
- A indirectly synchronizes with B (through X).
- A is sequenced before X which inter-thread happens before B
- A interthread happens before X and X interthread happens before B.
As you are not performing any of those cases there will be forms of your program where on some current hardware, it may fail.
In practice, the end of a time-slice will cause the memory to become coherent, or any form of barrier on the non-spinlock thread will ensure that the caches are flushed.
Not sure on the causes of the volatile read getting the "current value".
1.2 In practice, for known architectures and compilers, specifically for GCC and platforms it supports?
As the code is not consistent with the generalized CPU, from C++11 then it is likely this code will fail to perform with versions of C++ which try to adhere to the standard.
From cppreference : const volatile qualifiersVolatile access stops optimizations from moving work from before it to after it, and from after it to before it.
"This makes volatile objects suitable for communication with a signal handler, but not with another thread of execution"
So an implementation has to ensure that instructions are read from the memory location rather than any local copy. But it does not have to ensure that the volatile write is flushed through the caches to produce a coherent view across all the CPUs. In this sense, there is no time boundary on how long after a write into a volatile variable will become visible to another thread.
Also see kernel.org why volatile is nearly always wrong in kernel
Is this implementation safe on all architectures supported by GCC and Linux? (It is at least inefficient on some architectures, right?)
There is no guarantee the volatile message gets out of the thread which sets it. So not really safe. On linux it may be safe.
Is the while loop safe according to C++11 and its memory model?
No - as it doesn't create any of the inter-thread messaging primitives.