问题
There are a couple of questions on this site asking whether using a volatile variable for atomic / multithreaded access is possible: See here, here, or here for example.
Now, the C(++) standard conformant answer is obviously no.
However, on Windows & Visual C++ compiler, the situation seems not so clear.
I have recently answered and cited the official MSDN docs on volatile
Microsoft Specific
Objects declared as volatile are (...)
- A write to a volatile object (volatile write) has Release semantics; a reference to a global or static object? that occurs before a write to a volatile object in the instruction sequence will occur before that volatile write in the compiled binary.
- A read of a volatile object (volatile read) has Acquire semantics; a reference to a global or static object? that occurs after a read of volatile memory in the instruction sequence will occur after that volatile read in the compiled binary.
This allows volatile objects to be used for memory locks and releases in multithreaded applications.
[emphasis mine]
Now, reading this, it would appear to me that a volatile variable will be treated by the MS compiler as std::atomic would be in the upcoming C++11 standard.
However, in a comment to my answer, user Hans Passant wrote "That MSDN article is very unfortunate, it is dead wrong. You can't implement a lock with volatile, not even with Microsoft's version. (...)"
Please note: The example given in the MSDN seems pretty fishy, as you cannot generally implement a lock without atomic exchange. (As also pointed out by Alex.) This still leaves the question wrt. to the validity of the other infos given in this MSDN article, especially for use cases like here and here.)
Additionally, there are the docs for The Interlocked* functions, especially InterlockedExchange with takes a volatile(!?) variable and does an atomic read+write. (Note that one question we have on SO -- When should InterlockedExchange be used? -- does not authoritatively answer whether this function is needed for a read-only or write-only atomic access.)
What's more, the volatile docs quoted above somehow allude to "global or static object", where I would have thought that "real" acquire/release semantics should apply to all values.
Back to the question
On Windows, with Visual C++ (2005 - 2010), will declaring a (32bit? int?) variable as volatile allow for atomic reads and writes to this variable -- or not?
What is especially important to me is that this should hold (or not) on Windows/VC++ independently of the processor or platform the program runs on. (That is, does it matter whether it's a WinXP/32bit or a Windows 2008R2/64bit running on Itanum2?)
Please back up your answer with verifiable information, links, test-cases!
回答1:
Yes they are atomic on windows/vc++ (Assuming you meet alignment requirements etc or course)
However for a lock you would need an atomic test and set, or compare and exchange instuction or similar, not just an atomic update or read.
Otherwise there is no way to test the lock and claim it in one indivisable operation.
EDIT: As commented below, all aligned memory accesses on x86 of 32bit or below are atomic anyway. The key point is that volatile makes the memory accesses ordered. (Thanks for pointing this out in the comments)
回答2:
As of Visual C++ 2005 volatile variables are atomic. But this only applies to this specific class of compilers and to x86/AMD64 platforms. PowerPC for example may reorder memory reads/writes and would require read/write barriers. I'm not familar what the semantics are for gcc-class compilers, but in any case using volatile for atomics is not very portable.
reference, see first remark "Microsoft Specific": http://msdn.microsoft.com/en-us/library/12a04hfd%28VS.80%29.aspx
回答3:
A bit off-topic, but let's have a go anyway.
... there are the docs for The Interlocked* functions, especially InterlockedExchange which takes a volatile(!) variable ...
If you think about this:
void foo(int volatile*);
Does it say:
- the argument must be a pointer to a volatile int, or
- the argument may as well be a pointer to a volatile int?
The latter is the correct answer, since the function can be passed both pointers to volatile and non-volatile int's.
Hence, the fact that InterlockedExchangeX() has its argument volatile-qualified does not imply that it must operate on volatile integers only.
回答4:
The point is probably to allow stuff like
singleton& get_instance()
{
static volatile singleton* instance;
static mutex instance_mutex;
if (!instance)
{
raii_lock lock(instance_mutex);
if (!instance) instance = new singleton;
}
return *instance;
}
which would break if instance was written to before initialization was complete. With MSVC semantics, you are guaranteed that as soon as you see instance != 0, the object has finished being initialized (which is not the case without proper barrier semantics, even with traditional volatile semantics).
This double-checked lock (anti-)pattern is quite common actually, and broken if you don't provide barrier semantics. However, if there are guarantees that accesses to volatile variables are acquire + release barriers, then it works.
Don't rely on such custom semantics of volatile though. I suspect this has been introduced not to break existing codebases. In any way, don't write locks according to MSDN example. It probably doesn't work (I doubt you can write a lock using just a barrier: you need atomic operations -- CAS, TAS, etc -- for that).
The only portable way to write the double-checked lock pattern is to use C++0x, which provides a suitable memory model, and explicit barriers.
回答5:
under x86, these operations are guaranteed to be atomic without the need for LOCK based instructions such as Interlocked* (see intel's developer manuals 3A section 8.1):
basic memory operations will always be carried out atomically:
• Reading or writing a byte
• Reading or writing a word aligned on a 16-bit boundary
• Reading or writing a doubleword aligned on a 32-bit boundary
The Pentium processor (and newer processors since) guarantees that the following additional memory operations will always be carried out atomically:
• Reading or writing a quadword aligned on a 64-bit boundary
• 16-bit accesses to uncached memory locations that fit within a 32-bit data bus
The P6 family processors (and newer processors since) guarantee that the following additional memory operation will always be carried out atomically:
• Unaligned 16-, 32-, and 64-bit accesses to cached memory that fit within a cache line
This means volatile will only every serve to prevent caching and instruction reordering by the compiler (MSVC won't emit atomic operations for volatile variables, they need to be explicitly used).
来源：https://stackoverflow.com/questions/7007403/are-volatile-reads-and-writes-atomic-on-windowsvisualc