Instruction reordering on intel

Question

I'm trying to understand the instruction reordering by the following simple example:

int a;
int b;

void foo(){
   a = 1;
   b = 1;
}

void bar(){
   while(b == 0) continue;
   assert(a == 1);
}

It's known that in this example the assertion may fail if one thread executes foo, and another one executes bar. But I don't understand why. I consulted the Intel manual Vol. 3A, 8.2.2 and found the following:

Writes to memory are not reordered with other writes, with the following exceptions:

— streaming stores (writes) executed with the non-temporal move instructions (MOVNTI, MOVNTQ, MOVNTDQ, MOVNTPS, and MOVNTPD); and

— string operations (see Section 8.2.4.1).

There are no string operations here as well as I did not notice NT move instructions. So... Why is the reordering of writes possible?

Or does the memory matters in

Writes to memory are not reordered

? So when we have a and b cached and writes occur not to main memory, but to cache they can be.

Why do you think a and b will be stored in memory? There's no such requirement in C++ language, in corner case the compiler may generate code which will keep those two in registers only (for example). — Ped7g
No problem, searching the correct question is part of the process, and sometimes that process requires steps like this one. :) But the topic may be quite tricky, so you will have to be more precise to get some meaningful answer.. — Ped7g
you need prevent here compiler reordering. say by insert a = 1; _ReadWriteBarrier(); b = 1; on x86/x64 this will be enough — RbMm
Is this actually C or C++, or are you actually asking about assembly that does those operations in that order? Because obviously this is UB in C, and the C memory model applies for compile-time reordering and load hoisting, not the x86 memory model. IIRC, you did this in a previous question and wasted everyone's time sorting out C vs. x86 memory models, so please specify in the question that you actually mean if the asm looks like this. Obviously while(b == 0){} will become an infinite loop with any normal C compiler if it enters the loop at all. — Peter Cordes

Bathsheba Bathsheba · Accepted Answer · 2018-08-08T11:45:53

If one thread was running foo and another was running bar then the behaviour of your program would be undefined.

You are not allowed to make simultaneous read and writes on a non-atomic variable such as int.

So instruction reording is permissible in this instance.

Instruction reordering on intel

2 Answers