Do we have the guarantee that any atomic write will store the new value of the atomic variable in the main memory immediately?

Question

So, I was reading a lot about instruction and memory reordering and how we can prevent it, but i still have no answer to one qustion (probably because I'm not attentive enough). My question is: Do we have the guarantee that any atomic write will store the new value of the atomic variable in the main memory immediately? Let's take a look at a small example:

std::atomic<bool> x;
std::atomic<bool> y;
std::atomic<int> count;
void WritingValues()
{
   x.store(true, std::memory_order_relaxed);
   y.store(true, std::memory_order_relaxed);
}
void ReadValues()
{
   while( !y.load(std::memory_order_relaxed) );
   if( x.load(std::memory_order_relaxed) )
       ++count;
}
int main()
{
   x = false;
   y = false;
   count = 0;
   std::thread tA(WritingValues);
   std::thread tB(ReadValues);
   tA.join();
   tB.join();
   assert( count.load() != 0 );
}

So, here our assert can definitely fire, as we use std::memory_order_relaxed and do not prevent any instruction reordering (or memory reordering at compile time, i suppose that is the same thing). But if we place some compiler barrier in WritingValues to prevent instruction reordering , will everything be OK? I mean, does x.store(true, std::memory_order_relaxed) guarantees, that the write of that particular atomic variable will be directly into the memory, without any latency? Or does x.load(std::memory_order_relaxed) guarantess, that the value would be readed from the memory, not the cache with invalid value? In other words, this store guarantees only atomicity of the operation and have the same memory behaviour as usual non-atomic variable, or it also has influence on memory behaviour?

Answer 1

 I mean, does x.store(true, std::memory_order_relaxed) guarantees, that the  
 of that particular atomic variable will be directly into the memory,  
 without any latency?  

No it doesn't and in fact given bool and memory order relaxed there is no 'invalid' value if you read it only once, both true and false are ok.
Since relaxed memory order explicitly stays that no ordering is performed. Basically in your case it only means that after flipping from false to true, at some point it will become true for all other processes, but doesn't state 'whent' it will happen. So the only thing you can be sure here is that it won't become false again after becoming true. But there is no constraints on how long it will be false in another thread.
Also it guarantees that you won't see any partially written variable in another thread, but that's hardly the case for the bools.
You need to use aquire and release here. And even that won't give any guarantees about actual memory itself, only about program behavior, cache synchronization may do the trick even without bouncing data back and froth to the memory.

Answer 2

Since all of the load and store instructions are atomic they each a single machine instruction so the two threads never "interrupt each other" in the "middle" of a load or store instruction.

The title of your question is "Do we have the guarantee that any atomic write will store the new value of the atomic variable in the main memory immediately?". But the very definition of an atomic instruction is that it cannot be interrupted by a context switch, hardware interrupt, software expection -- nothing!

std::memory_order_relaxed allows for some reordering of instructions in a single function. See for example this question. It is almost the same as your question but you have memory_order_relaxed in ReadValues() instead of memory_order_acquire . In this function it is possible that the spinlock on variable y is placed after the counter increment due to the relaxed condition (compiler reordering etc.). In any case the ASSERT may fail because y may be set to true before x is in WriteValues() due to the memory reordering allowed by memory_order_relaxed (referencing the answer in the similar question.