Mallocator and 'shared'

[ag0aep6g via Digitalmars-d-learn]

On 02/13/2017 01:27 AM, Moritz Maxeiner wrote:
> __gshared int a = 0;
>
> // thread 1:
> a = 1;
> int b = a;
> writeln(b);
>
> // thread 2:
> a += 1;
>
> In the above, you may expect `b` to be either 1, or 2, depending on how
> the cpu interleaves the memory access, but it can, in fact, also be 0,
> since neither the compiler, nor the cpu can detect any reason as to why
> `a = 1` should need to come before `int b = a` and may thus reorder the
> write and the read.

This doesn't make sense to me. b depends on a. If I run thread 1 alone, 
I can expect b to be 1, no?  Thread 2 can then a) read 0, write 1; or b) 
read 1, write 2. How can b be 0 when the writeln is executed?

An example like this makes more sense to me:

----
shared int a = 0, b = 0;

// Thread 1:
a = 1;
b = 2;

// Thread 2:
writeln(a + b);
----

One might think that thread 2 cannot print "2", because from the order 
of statements the numbers must be 0 + 0 or 1 + 0 or 1 + 2. But thread 1 
might execute `b = 2` before `a = 1`, because the order doesn't matter 
to thread 1. So 0 + 2 is possible, too.