.NET memory model, volatile variables, and test-and-set: what is guaranteed?

Question

I know that the .NET memory model (on the .NET Framework; not compact/micro/silverlight/mono/xna/what-have-you) guaranteed that for certain types (most notably primitive integers and references) operations were guaranteed to be atomic.

Further, I believe that the x86/x64 test-and-set instruction (and Interlocked.CompareExchange) actually references the global memory location, so if it succeeds another Interlocked.CompareExchange would see the new value.

Finally, I believe that the volatile keyword is an instruction to the compiler to propagate reads & writes ASAP and to not reorder operations concerning this variable (right?).

This leads to a few questions:

1. Are my beliefs above correct?
2. Interlocked.Read does not have an overload for int, only for longs (which are 2 WORDs and thus are not normally read atomically). I always assumed that the .NET memory model guaranteed that the newest value would be seen when reading ints/references, however with processor caches, registers, etc. I'm starting to see this may not be possible. So is there a way to force the variable to be re-fetched?
3. Is volatile sufficient to solve the above problem for integers and references?
4. On x86/x64 can I assume that...

If there are two global integer variables x and y, both initialized to 0 that if I write:

x = 1;
y = 2;

That NO thread will see x = 0 and y = 2 (i.e. the writes will occur in order). Does this change if they are volatile?

Answer 1

No, the volatile keyword and the atomicity guarantee are much too weak. You need a memory barrier to ensure that. You can get one explicitly with the Thread.MemoryBarrier() method.

Answer 2

• Only reads and writes to variables that are at most 32 bits wide (and 64 bits wide on x64 systems) are atomic. All this means is that you won't read an int and get a half-written value. It doesn't mean arithmetic is atomic.
• The interlocked operations also act as memory barriers, so yes, Interlocked.CompareExchange will see the updated value.
• See this page. Volatile does not mean ordered. Some compilers may choose not to re-order operations on volatile variables, but the CPU is free to re-order. If you want to stop the CPU from re-ordering instructions, use a (full) memory barrier.
• The memory model ensures that reads and writes are atomic, and using the volatile keyword ensures that reads will always come from memory, not from a register. So you will see the newest value. This is because x86 CPUs will invalidate the cache when appropriate - see this and this. Also, see InterlockedCompareExchange64 for how to atomically read 64-bit values.
• And finally, the last question. The answer is a thread could in fact see x = 0 and y = 2, and using the volatile keyword doesn't change that because the CPU is free to re-order instructions. You need a memory barrier.

Summary:

1. The compiler is free to re-order instructions.
2. The CPU is free to re-order instructions.
3. Word-sized reads and writes are atomic. Arithmetic and other operations are not atomic because they involve a read, compute, then write.
4. Word-sized reads from memory will always retrieve the newest value. But most of the time you don't know if you're actually reading from memory.
5. A full memory barrier stops (1) and (2). Most compilers allow you to stop (1) by itself.
6. The volatile keyword ensures you're reading from memory - (4).
7. The interlocked operations (the lock prefix) allow multiple operations to be atomic. For example, a read + write (InterlockedExchange). Or a read + compare + write (InterlockedCompareExchange). They also act as memory barriers, so (1) and (2) are stopped. They always write to memory (obviously), so (4) is ensured.

Answer 3

Came across this old thread. The answers from Hans and wj32 are all correct except for the part regarding volatile.

Specifically regarding your question

On x86/x64 can I assume that... If there are two global integer variables x and y, both initialized to 0 that if I write: x = 1; y = 2;

That NO thread will see x = 0 and y = 2 (i.e. the writes will occur in order). Does this change if they are volatile?

If y is volatile, the write to x is guarantee to happen before the write to y, therefore no thread will ever see x = 0 and y = 2. That is because the write to a volatile variable has the "release semantic" (logically equivalent to the emission of a release fence), i.e. all read/write instructions before it won't move pass it. (This implies that if x is volatile but y is not, you might still see the unexpected x = 0 and y = 2.) See the description & code example in the C# spec for more details.